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10.1

Learning Objectives

In this unit, the student is exposed to

• oscillatory motion – periodic motion and non-
• simple harmonic motion
• angular harmonic motion
• linear harmonic oscillator – both horizontal an
• combination of springs – series and parallel
• simple pendulum
• expression of energy – potential energy, kinetic
• graphical representation of simple harmonic m
• types of oscillation – free, damped, maintained
• concept of resonance

INTRODUCTION

Have you seen the Thanjavur Dancing Doll (In Tamil, it is called ‘Thanjavur thalayatti bommai’)?. It is a world famous Indian cultural doll (image 10.1). What does this _Life is a constant oscillation between

image 10.1. Thanjavur dancing doll

doll do when disturbed? It will dance such that the head and body move continuously in a to and fro motion, until the movement gradually stops. Similarly, when we walk on the road, our hands and legs will move front and back. Again similarly, when a mother swings a cradle to make her child sleep, the cradle is made to move in to and fro motion. All these motions are different from the motion that we have discussed so far. These motions are shown in image 10.2. Generally, they are known as oscillatory motion or vibratory motion. A similar motion occurs even at atomic levels.

When the temperature is raised, the atoms in a solid vibrate about their mean position or equilibrium position. The study of vibrational motion is very important in engineering applications, such as, designing the structure of building, mechanical equipments, etc.

Periodic and non- periodic motion

Motion in physics can be classified as repetitive (periodic motion) and non- repetitive (non-periodic motion).

1. Periodic motion which repeats itself in a fixed time interval is known as periodic motion.

Examples : Hands in pendulum clock, swing of a cradle, the revolution of the Earth around the Sun, waxing and waning of Moon, etc.

2. Non-Periodic motion** Any motion which does not repeat itself after a regular interval of time is known as non-periodic motion. Example : Occurance of Earth quake, eruption of volcano, etc.

EXAMPLE 10.1

Classify the following motions as periodic and non-periodic motions?.

a. Motion of Halley’s comet.

b. Motion of clouds.

c. Moon revolving around the Earth.

Solution

a. Periodic motion

b. Non-periodic motion c. Periodic motion

EXAMPLE 10.2

Which of the following functions of time represent periodic and non-periodic motion?. a. sin ωt + cos ωt b. ln ωt

Solution

a. Periodic

b. Non-periodic

Question to ponder

Discuss “what will happen if the motion of the Earth around the Sun is not a periodic motion”.

Oscillatory motion

When an object or a particle moves back and forth repeatedly for some duration of time its motion is said to be oscillatory (or vibratory). Examples; our heart beat, swinging motion of the wings of an insect, grandfather’s clock (pendulum clock), etc. Note that all oscillatory motion are periodic whereas all periodic motions need not be oscillation in nature. see image 10.3

|note

A simple harmonic motion is a special type of oscillatory motion. But all oscillatory motions need not be simple harmonic .

image 10.3 Oscillatory or vibratory motion

SIMPLE HARMONIC MOTION (SHM)

image 10.4 Simple Harmonic Motion

Equilibrium position

Note

Simple harmonic motion is a special type of oscillatory motion in which the acceleration or force on the particle is directly proportional to its displacement from a fixed point and is always directed towards that fixed point. In one dimensional case, let x be the displacement of the particle and ax be the acceleration of the particle, then

$$ a_x \propto x (10.1) $$ $$ a_x = -b , x (10.2) $$

where b is a constant which measures acceleration per unit displacement and dimensionally it is equal to T −2. By multiplying by mass of the particle on both sides of equation (10.2) and from Newton’s second law, the force is

$$F_x = -k x(10.3)$$

where k is a force constant which is defined as force per unit length. The negative sign indicates that displacement and force (or acceleration) are in opposite directions. This means that when the displacement of the particle is taken towards right of equilibrium position (x takes positive value), the force (or acceleration) will point towards equilibrium (towards left) and similarly,

when the displacement of the particle is taken towards left of equilibrium position (x takes negative value), the force (or acceleration) will point towards equilibrium (towards right). This type of force is known as restoring force because it always directs the particle executing simple harmonic motion to restore to its original (equilibrium or mean) position. This force (restoring force) is central and attractive whose center of attraction is the equilibrium position.

In order to represent in two or three dimensions, we can write using vector notation

$$F = -k \vec{r}(10.4)$$

where  r is the displacement of the particle from the chosen origin. Note that the force and displacement have a linear relationship. This means that the exponent of force  F and the exponent of displacement  r are unity. The sketch between cause (magnitude of force |  F |) and effect (magnitude of displacement |  r |) is a straight line passing through second and fourth quadrant as shown in

The projection of uniform circular motion on a diameter of SHM

Consider a particle of mass m moving with uniform speed v along the circumference of a circle whose radius is r in anti-clockwise direction (as shown in image 10.6). Let us assume that the origin of the coordinate system coincides with the center O of the circle. If ω is the angular velocity of the particle and θ the angular displacement of the particle at any instant of time t, then θ = ωt. By projecting the uniform circular motion on its diameter gives a simple harmonic motion. This means that we can associate a map (or a relationship) between uniform circular (or revolution) motion to vibratory motion. Conversely, any vibratory motion or revolution can be mapped to uniform circular motion. In other words, these two motions are similar in nature.

Let us first project the position of a particle moving on a circle, on to its vertical diameter or on to a line parallel to vertical diameter as shown in image 10.7. Similarly, we can do it for horizontal axis or a line parallel to horizontal axis.

As a specific example, consider a spring mass system (or oscillation of pendulum) as shown in image 10.8. When the spring moves up and down (or pendulum moves to and fro), the motion of the mass or bob is mapped to points on the circular motion.

image 10.8 Motion of spring mass (or simple pendulum) related to uniform circular motion

Thus, if a particle undergoes uniform circular motion then the projection of the particle on the diameter of the circle (or on a line parallel to the diameter ) traces straightline motion which is simple harmonic in nature. The circle is known as reference circle of the simple harmonic motion. _The simple harmonic motion can also be defined as the motion of the projection of a particle on any diameter of a circle of reference.

Activity

a. Sketch the projection of spiral in motion as a wave form.

b. Sketch the projection of spiral out motion as a wave form.

Displacement, velocity, acceleration and its graphical representation - SHM

The distance travelled by the vibrating particle at any instant of time t from its mean position is known as displacement.

Let P be the position of the particle on a circle of radius A at some instant of time t as shown in image 10.9. Then its displacement y at that instant of time t can be derived as follows In ∆_OPN_

sin sinθ θ= ⇒ = ON OP

$$\sin{\theta}=\frac{ON}{OP} \Rightarrow ON = OP\sin{\theta}\quad(10.5)$$

sin s θ θ = ⇒ = in ON = OPON OP (10.5)

$$y = A\sin\omegat(10.6) $$

The displacement y takes maximum value (which is equal to A) when sin ωt = 1. This maximum displacement from the mean position is known as amplitude (A) of the vibrating particle. For simple harmonic motion, the amplitude is constant. But, in general, for any motion other than simple harmonic, the amplitude need not be constant, it may vary with time.

Velocity

The rate of change of displacement is velocity. Taking derivative of equation (10.6) with respect to time, we get

v dy dt

d dt

= = (A sin ωt)

For circular motion (of constant radius), amplitude A is a constant and further, for uniform circular motion, angular velocity ω is a constant. Therefore,

v dy dt $$y = A\sin(\omega t + \phi)(10.6)$$

Using trigonometry identity, sin2 ωt + cos2 ωt =1 ⇒ cos ωt = 1 2-sin ω_t_ we get v A t= −ω ω1 2sin

From equation (10.8), when the displacement y = 0, the velocity v = ωA (maximum) and for the maximum displacement y = A, the velocity v = 0 (minimum). As displacement increases from zero to maximum, the velocity decreases from maximum to zero. This is repeated. Since velocity is a vector quantity, equation (10.7) can also be deduced by resolving in to components.

Acceleration

The rate of change of velocity is acceleration. Taking derivative of equation 10.7 with respect to time,

a dv dt

d dt

A t= = ( )ω ωcos

$$ya = -{\omega}^2 A \sin{\omega t} = -{\omega}^2 y(10.9)$$

∴

$$a = -{\omega}^2 A\sin{\omega t}= -{\omega}^2 y(10.10)$$

(y = 0), velocity of the particle is maximum but the acceleration of the particle is zero. At the extreme position (y = ±_A_), the velocity of the particle is zero but the acceleration is maximum _Aω_2 acting in the opposite direction.

EXAMPLE 10.3

Which of the following represent simple harmonic motion? (i) x = A sin ω_t_ + B cos ωt (ii) x = A sin ωt+ B cos 2_ωt_ (iii) x = A _e_iωt

(iv) x = A ln ωt

Solution

(i)$$[ x = A \sin(\omega t) + B \cos(\omega t) ]$$

This differential equation is similar to the differential equation of SHM (equation 10.10). Therefore, x = A sin ωt + B cos ωt represents SHM. (ii) x =A sin ωt + B cos2_ωt_

$$[ \frac{d^2 x}{dt^2} = -A \omega^2 e^{i \omega t} \quad (\text{since } i^2 = -1) ]$$

$$ [ \frac{dx}{dt} = A \omega \cos(\omega t) - B \omega \sin(\omega t) ]$$

$$[ \frac{d^2 x}{dt^2} = -\omega^2 (A \sin(\omega t) + B \cos(\omega t)) ]$$

$$[ \frac{d^2 x}{dt^2} = -\omega^2 x ]$$

This differential equation is not like the differential equation of a SHM (equation 10.10).

Therefore, x = A sin ωt + B cos 2ωt does not represent SHM. (iii) $$ [ x = A \ln(\omega t) ]$$

$$ [ \frac{dx}{dt} = \frac{A}{\omega t} = \frac{A}{t} ] $$

$$ [ \frac{d^2 x}{dt^2} = -\frac{A}{t^2} \Rightarrow \frac{d^2 x}{dt^2} = -\omega^2 x ] $$

This differential equation is like the differential equation of SHM (equation 10.10).

Therefore, x = A eiωt represents SHM.

(iv) $$ [ x = A \ln(\omega t) ] $$

$$[ \frac{dx}{dt} = \left( \frac{A}{\omega t} \right) \omega = \frac{A}{t} ] $$

$$ [ \frac{d^2 x}{dt^2} = -\frac{A}{t^2} \Rightarrow x’’ = -\omega^2 x ]$$

This differential equation is not like the differential equation of a SHM (equation 10.10). Therefore, x = A ln ωt does not represent SHM.

EXAMPLE 10.4

Consider a particle undergoing simple harmonic motion. The velocity of the particle at position _x_1 is _v_1 and velocity of the particle at position _x_2 is _v_2. Show that the ratio of time period and amplitude is

$$[ \frac{T}{A} = 2\pi \sqrt{\frac{x_2^2 - x_12}{v_12 x_2^2 - v_2^2 x_1^2}} ]$$

| image 10.10 Variation of displacement, velocity and acceleration at dierent instant of time |

Solution

Using equation (10.8)

$$[ v = \omega \sqrt{A^2 - x^2} \Rightarrow v^2 = \omega^2 (A^2 - x^2) ]$$

Therefore, at position ( x_1 ):

$$[ v_1^2 = \omega^2 (A^2 - x_1^2) \quad (1) ]$$

Similarly, at position ( x_2 ):

$$[ v_2^2 = \omega^2 (A^2 - x_2^2) \quad (2) ]$$

Subtracting (2) from (1), we get:

$$[ v_1^2 - v_2^2 = \omega^2 (x_2^2 - x_1^2) ]$$

$$[ \omega = \sqrt{\frac{v_1^2 - v_22}{x_22 - x_1^2}} ]$$

Time period, frequency, phase, phase difference and epoch in SHM.

(i) Time period The time period is defined as the time taken by a particle to complete one oscillation. It is usually denoted by T. For one complete revolution, the time taken is t = T, therefore

$$[ \omega(T = 2\pi) \Rightarrow T = \frac{2\pi}{\omega} \quad (10.11) ]$$

Then, the displacement of a particle executing simple harmonic motion can be written either as sine function or cosine function.

$$[ y(t) = A \sin\left(\frac{2\pi}{T}t\right) ]$$

$$[ y(t + T) = A \sin\left(\frac{2\pi}{T}(t + T)\right) ]$$

$$ [ y(t + T) = A \sin\left(\frac{2\pi}{T}t + 2\pi\right) = A \sin\left(\frac{2\pi}{T}t\right) = y(t) ] $$

y(t + T) = y(t) Thus, the function repeats after one time period. This y(t) is an example of periodic function.

(ii) Frequency and angular frequency

The number of oscillations produced by the particle per second is called frequency. It is denoted by f. SI unit for frequency is _s_−1 or hertz (In symbol, Hz). Mathematically, frequency is related to time period by

$$ f = \frac{1}{T}(10.12)$$

The number of cycles (or revolutions) per second is called angular frequency. It is usually denoted by the Greek small letter ‘omega’, ω. Comparing equation (10.11) and equation (10.12), angular frequency and frequency are related by $$v = 2\pi f(10.13)$$

SI unit for angular frequency is rad _s_−1. (read it as radian per second)

(iii) Phase

The phase of a vibrating particle at any instant completely specifies the state of the particle. It expresses the position and direction of motion of the particle at that instant with respect to its mean position (image 10.11).

$$y = A \sin(\omega t + \phi_0)(10.14)$$

where ω_t_ + φ0 = φ is called the phase of the vibrating particle. At time t = 0 s (initial time), the phase φ = φ0 is called epoch (initial phase) where φ0 is called the angle of epoch. Phase difference: Consider two particles executing simple harmonic motions. Their equations are y_1 = A sin(ω_t + φ1) and y2 = A sin(ωt + φ2), then the phase difference ∆φ= (ω_t_ + φ2) − (ω_t_ + φ1) = φ2 −φ1.

image 10.11 The phase of vibrating particle at two instant of time.

EXAMPLE 10.5

A nurse measured the average heart beats of a patient and reported to the doctor in terms of time period as 0.8 s. Express the heart beat of the patient in terms of number of beats measured per minute.

Solution

Let the number of heart beats measured be f. Since the time period is inversely proportional to the heart beat, then

$$f = \frac{1}{T} = \frac{1}{0.8} = 1.25 , \text{s}^{-1} $$

One minute is 60 second,

(1 second = 1 60

minute ⇒ 1 s−1 = 60 min−1)

f =1.25 s−1 ⇒ f = 1.25 × 60 min−1 = 75 beats per minute

EXAMPLE 10.6

Calculate the amplitude, angular frequency, frequency, time period and initial phase for the simple harmonic oscillation given below

a. y = 0.3 sin (40πt + 1.1)

b. y = 2 cos (πt)

c. y = 3 sin (2πt − 1.5)

Solution

Simple harmonic oscillation equation is y = A sin(ω_t_ + φ0) or y =A cos(ω_t_ + φ0) a. For the wave, y = 0.3 sin(40π_t_ +1.1)

Amplitude is A = 0.3 unit

Angular frequency ω = 40π rad s−1

Frequency f Hz= = = ω π

$$f = \frac{\theta}{2\pi} = \frac{40\pi}{2\pi} = 20 , \text{Hz} $$

$$f = \frac{\theta}{2\pi} = \frac{40\pi}{2\pi} = 20 , \text{Hz} $$

$$\phi_0 = 1.1 , \text{rad} $$

Initial phase is φ0 = 1.1 rad

b. For the wave, y = 2 cos (π_t_)

Amplitude is A = 2 unit

Angular frequency ω = π rad s−1

Frequency f Hz= = = ω π

$$f = \frac{\omega}{2\pi} = \frac{\pi}{2\pi} = 0.5 , \text{Hz} $$

$$T = \frac{1}{f} = \frac{1}{0.5} = 2 , \text{s} $$

$$\phi_0 = 0 , \text{rad} $$ c. For the wave, y = 3 sin(2πt + 1.5)

Amplitude is A = 3 unit

Angular frequency ω = 2π rad s−1

$f = \frac{\omega}{2\pi} = \frac{2\pi}{2\pi} = 1 , \text{Hz} $$

$$T = \frac{1}{f} = \frac{1}{1} = 1 , \text{s} $$

$$\phi_0 = 1.5 , \text{rad} $$ Initial phase is φ0 = 1.5 rad

EXAMPLE 10.7

Show that for a simple harmonic motion, the phase difference between

a. displacement and velocity is π/2

2 radianor 90°.

b. velocity and acceleration is π/2 radian

or 90°.

c. displacement and acceleration is π radian or 180°.

Solution

a. The displacement of the particle executing simple harmonic motion

y = A sinω_t_

Velocity of the particle is

$$v = A(\theta) \cos(\omega t) = -A(\theta) \sin(\omega t + \frac{\pi}{2}) $$

The phase difference between displacement and velocity is π/2 .

The phase difference between displacement and velocity is π/2 .

b. The velocity of the particle is

$$v = A \omega \cos(\omega t)$$

Acceleration of the particle is

$$a = -A \omega^{2} \sin(\omega t) = A \omega^{2} \cos\left(\omega t + \frac{\pi}{2}\right) $$

2 sin cos t

The phase difference between velocity and acceleration is π

2 .

c. The displacement of the particle is y = A sinω_t_

Acceleration of the particle is

$$a = -A \omega^{2} \sin(\alpha) = A \omega^{2} \sin(\alpha + \pi) $$

The phase difference between displacement and acceleration is π radian.

ANGULAR SIMPLE HARMONIC MOTION

Time period and frequency of angular SHM

When a body is allowed to rotate freely about a given axis then the oscillation is known as the angular oscillation. The point

at which the resultant torque acting on the body is taken to be zero is called mean position. If the body is displaced from the mean position, then the resultant torque acts such that it is proportional to the angular displacement and this torque has a tendency to bring the body towards the mean position. (Note: Torque is explained in unit 5) Let

θ be the angular displacement of the body and the resultant torque τ acting on the body is

$$\bar{\tau} \propto \bar{\theta} \quad \text{(10.15)} (10.15)$$

$$(\bar{\tau} = -k \dot{\theta} \quad \text)(10.16)$$

κ is the restoring torsion constant, which is torque per unit angular displacement. If I is the moment of inertia of the body and α is the angular acceleration then

$$\tau = I \alpha = k \theta $$

$$\frac{d^2\theta}{dt^2} = -\frac{K}{I}\theta \quad (10.17) $$

This differential equation resembles simple harmonic differential equation. So, comparing equation (10.17) with simple harmonic motion given in equation (10.10), we have

$$\theta = \sqrt{\frac{K}{I}} \ \text{rads}^{-1} (10.18)$$

The frequency of the angular harmonic motion (from equation 10.13) is

$$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \quad \text{Hz} (10.19)$$

The time period (from equation 10.12) is

T I =2π

$$T = 2\pi \sqrt{\frac{I}{k}} \ \text{second} (10.20)$$

LINEAR SIMPLE HARMONIC OSCILLATOR (LHO)

Horizontal oscillations of a spring-mass system

image 10.13 Horizontal oscillation of a spring-mass system

Consider a system containing a block of mass m attached to a massless spring with stiffness constant or force constant or spring constant k placed on a smooth horizontal surface (frictionless surface) as shown in image 10.13. Let _x_0 be the equilibrium position or mean position of mass m when it is left undisturbed. Suppose the mass is displaced through a small displacement x towards right from its equilibrium position and then released, it will oscillate back and forth about its mean position _x_0. Let F be the restoring force (due to stretching of the spring) which is proportional to the amount of displacement of block. For

one dimensional motion, mathematically, we have

$$F \propto x \F = -kx$$

where negative sign implies that the restoring force will always act opposite to the direction of the displacement. This equation is called Hooke’s law (refer to unit 7). Notice that, the restoring force is linear with the displacement (i.e., the exponent of force and displacement are unity). This is not always true; in case if we apply a very large stretching force, then the amplitude of oscillations becomes very large (which means, force is proportional to displacement containing higher powers of x) and therefore, the oscillation of the system is not linear and hence, it is called non-linear oscillation. We restrict ourselves only to linear oscillations throughout our discussions, which means Hooke’s law is valid (force and displacement have a linear relationship). From Newton’s second law, we can write the equation for the particle executing simple harmonic motion

$$m \frac{d^2 x}{dt^2} = -kx$$

$$\frac{d^2 x}{dt^2} = -\frac{k}{m}x \quad (10.21)(10.21)$$

Comparing the equation (10.21) with simple harmonic motion equation (10.10), we get

$$\dot{\Theta} = \frac{k}{m}$$

which means the angular frequency or natural frequency of the oscillator is

$$\omega = \sqrt{\frac{k}{m}} \ \text{rad s}^{-1} $$

rad _s_−1 (10.22) The frequency of the oscillation is

and the time period of the oscillation is

$$f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \quad \text{Hertz} $$

$$T = \frac{1}{f} = 2\pi \sqrt{\frac{m}{k}} \text{ seconds}(10.24) $$

Notice that in simple harmonic motion, the time period of oscillation is independent of amplitude. This is valid only if the amplitude of oscillation is small. The solution of the differential equation of a SHM may be written as

$$x(t) = A \cos(\omega t + \phi) (10.25)$$

Or

$$x(t) = A \cos(\omega t + \phi) (10.26)$$

where A, ω and ϕ are constants. General solution for differential equation 10.21 is x(t) = A sin(ω_t_ +φ)+ B cos(ω_t_ +φ) where A and B are contants.

(a) Since, mass is inertial property and spring constant is an elastic property,

$$\text{Time period is, } T = 2\pi \sqrt{\frac{\text{Inertial property}}{\text{Elastic property}}} = 2\pi \sqrt{\frac{m}{k}}$$

Note

Τ= =2 2π π Inertial property Elasticproperty

displacement acceleration

(b) Displacement acceleration $$ \quad & \frac{\text{Displacement}}{\text{acceleration}} = \frac{x}{\frac{d^2x}{dt^2}} = \frac{m}{k}$$ & \text{modulus value or magnitude is } \left| \frac{m}{k} \right| \ & \text{hence, time period } T = 2\pi\sqrt{\left|\frac{m}{k}\right|} \end{align*}$$

Let us consider a massless spring with stiffness constant or force constant k attached to a ceiling as shown in image 10.15. Let the length of the spring before loading mass m be L. If the block of mass m is attached to the other end of spring, then the spring elongates by a length l. Let F1 be the restoring force due to stretching of spring. Due to mass m, the gravitational force acts vertically downward. We can draw free-body diagram for this system as shown in image 10.15. When the system is under equilibrium,

$$F_1 - mg = 0 \quad (10.27)$$

But the spring elongates by small displacement l, therefore,

$$F_c \propto l \Rightarrow F_c = -kl \quad (10.28)$$

Substituting equation (10.28) in equation (10.27), we get

$$-k l + mg = 0 \tag{10.27}$$

or

$$\frac{m}{k} = \frac{l}{g} \tag(10.29)$$

Suppose we apply a very small external force on the mass such that the mass further displaces downward by a displacement y, then it will oscillate up and down. Now, the restoring force due to this stretching of spring (total extension of spring is y + l ) is

$$F_2 = -k(y + D) = -ky - kD \quad (10.30) (10.30)$$

Since, the mass moves up and down with acceleration d y dt 2 2 , by drawing the free body diagram for this case, we get

$$-kv - kt + mg = m \frac{d^2 y}{dt^2} \quad (10.31)$$

The net force acting on the mass due to this stretching is

$$ ( F = F_2 + mg ) ( F = -ky - k_1 y + mg ) \quad (10.32)$$

The gravitational force opposes the restoring force. Substituting equation (10.29) in equation (10.32), we get

F = -ky - k_1 y + mg

Applying Newton’s law, we get

$$[ m \frac{d^2 y}{dt^2} = -ky ]$$

$$[ \frac{d^2 y}{dt^2} = -\frac{k}{m}y \quad (10.33) ]$$

The above equation is in the form of simple harmonic differential equation. Therefore, we get the time period as

T m k

$$ [T = 2\pi \sqrt{\frac{m}{k}} \text{ second} ]$$

Note

The time period obtained for horizontal oscillations of spring and for verticaloscillations of spring are found to be equal.

The time period can be rewritten using equation (10.29)

$$ [ T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{l}{g}} \quad \text{(10.35)} ]$$

The acceleration due to gravity g can be computed from the formula $$ [ g = 4\pi^2 \left( \frac{l}{T^2} \right) , \text{m} , \text{s}^{-2} \quad \text{(10.36)} ]$$

EXAMPLE 10.8

A spring balance has a scale which ranges from 0 to 25 kg and the length of the scale is 0.25m. It is taken to an unknown planet X where the acceleration due to gravity is 11.5 m s−2. Suppose a body of mass M kg is suspended in this spring and made to oscillate with a period of 0.50 s. Compute the gravitational force acting on the body.

Solution

Let us first calculate the stiffness constant of the spring balance by using equation (10.29),

$$ [ k = \frac{mg}{l} = \frac{25 \times 11.5}{0.25} = 1150 , \text{N/m}^{-1} ]$$

The time period of oscillations is given by

$$[ T = 2\pi \sqrt{\frac{M}{k}} ]$$

, where M is the mass of the

body. Since, M is unknown, rearranging, we get $$[ M = \frac{kT2}{4\pi2} = \frac{(1150)(0.5)2}{4\pi2} = 7.3 , \text{kg} ]$$

The gravitational force acting on the body is W = Mg = 7.3 × 11.5 = 83.95 N ≈ 84 N

Combinations of springs

image 10.16 Combination of spring as a shock-absorber in the motor cycle

Spring constant or force constant, also called as stiffness constant, is a measure of the stiffness of the spring.

Larger the value of the spring constant, stiffer is the spring. This implies that we need to apply more force to compress or elongate the spring. Similarly, smaller the value of spring constant, the spring can be stretched (elongated) or compressed with lesser force.

Springs can be connected in two ways. Either the springs can be connected end to end, also known as series connection, or alternatively, connected in parallel. In the following subsection, we compute the effective spring constant when

a. Springs are connected in series

b. Springs are connected in parallel

a. Springs connected in series When two or more springs are connected in series, we can replace (by

removing) all the springs in series with an equivalent spring (effective spring) whose net effect is the same as if all the springs are in series connection. Given the value of individual spring constants _k_1, _k_2, _k_3,… (known quantity), we can establish a mathematical relationship to find out an effective (or equivalent) spring constant ks (unknown quantity). For simplicity, let us consider only two springs whose spring constant are _k_1 and _k_2 and which can be attached to a mass m as shown in image 10.17. The results thus obtained can be generalized for any number of springs in series.

Let F be the applied force towards right as shown in image 10.18. Since the spring constants for different spring are different and the connection points between them is not rigidly fixed, the strings can stretch in different lengths. Let x1 and x2 be the elongation of springs from their equilibrium position (un-stretched position) due to the applied force F. Then, the net displacement of the mass point is

$$[ x = x_1 + x_2 \quad \text{(10.37)} ]$$

From Hooke’s law, the net force

$$ [ F = -k_s (x_1 + x_2) \Rightarrow x_1 + x_2 = -\frac{F}{k_s} \quad \text{(10.38)} ]

$$

For springs in series connection −_k_1_x_1 = −_k_2_x_2 = F

⇒ $$ [ x_1 = -\frac{F}{k_1} \quad \text{and} \quad x_2 = -\frac{F}{k_2} \quad \text{(10.39)} ]$$

Therefore, substituting equation (10.39) in equation (10.38), the effective spring constant can be calculated as

$$[ \frac{F}{k_1} = \frac{F}{k_2} = \frac{F}{k_3} ]$$

$$[ \frac{1}{k_s} = \frac{1}{k_1} + \frac{1}{k_2} ]$$

or

$$ [ k_s = \frac{k_1 k_2}{k_1 + k_2} \quad \text{Nm}^{-1} \quad \text{(10.40)} ] $$

Suppose we have “n” springs connected in series, the effective spring constant in series is

$$[ \frac{1}{k} + \frac{1}{k_1} + \frac{1}{k_2} + \ldots + \frac{1}{k_n} = \frac{1}{k} \sum_{j=1}^{n} \frac{1}{k_j} \quad \text{(10.41)} ]$$

If all spring constants are identical i.e., _k_1 = _k_2 =… = _k_n = k then

$$[ k_2 = \ldots = k_n = k \text{ then} ]$$

$$[ \frac{1}{k} = \frac{n}{k} \implies k = \frac{n}{1} \quad \text{(10.42)} ]$$

This means that the effective spring constant reduces by the factor “n”. Hence, for springs in series connection, the effective spring constant is lesser than the individual spring constants. From equation (10.39), we have,

$$[ k_1 x_1 = k_2 x_2 ]$$

Then the ratio of compressed distance or elongated distance _x_1 and _x_2 is

$$[ \frac{x_2}{x_1} = \frac{k_1}{k_2} \quad \text{(10.43)} ]$$ The elastic potential energy stored in first and second springs are $$[ U_1 = \frac{1}{2} k_1 x_1^2 ]$$

21 2

= and

$$[ U_1 = \frac{1}{2} k_1 x_1^2 ]$$

$$[ U_2 = \frac{1}{2} k_2 x_2^2 ]$$

$$respectively. Then, their ratio is

$$[ \frac{U_1}{U_2} = \frac{\frac{1}{2}k_1x_12}{\frac{1}{2}k_2x_22} = \left(\frac{k_1}{k_2}\right)^2 \left(\frac{x_1}{x_2}\right)^2 = \left(\frac{k_1x_1}{k_2x_2}\right)^2 \quad \text{(10.44)} ]$$ = (10.44)

Note

The reciprocal of stiffness constant is called flexibility constant or compliance,

denoted by C. It is measured in m N^-1

If n springs are connected in series :

net compliance = $$[ e C_s = \sum_{i=1}^{n} C_i ]$$

If n springs are connected in parallel : $$[ \frac{1}{C_p} = \sum_{i=1}^{n} \frac{1}{C_i} ]$$

EXAMPLE 10.9

Consider two springs whose force constants are 1 N m−1 and 2 N m−1 which are connected in series. Calculate the effective spring constant (ks ) and comment on ks .

Solution $$[ \frac{1}{C_p} = \sum_{i=1}^{n} \frac{1}{C_i} ]$$

Therefore, the effective spring constant is lesser than both _k_1 and _k_2.

b. Springs connected in parallel When two or more springs are connected in parallel, we can replace (by removing) all these springs with an equivalent spring (effective spring) whose net effect is same as if all the springs are in parallel connection. Given the values of individual spring constants to be _k_1,_k_2,_k_3, … (known quantities), we can establish a mathematical relationship to find out an effective (or equivalent) spring constant _k_p (unknown quantity). For simplicity, let us consider only two springs of spring constants _k_1and _k_2 attached to a mass m as shown in image 10.19. The results can be generalized to any number of springs in parallel.

Let the force F be applied towards right as shown in image 10.20. In this case, both the springs elongate or compress by the same amount of displacement. Therefore, net force for the displacement of mass m is

$$[ F = -kx \quad \text{(10.45)} ]$$

where kp is called effective spring constant. Let the first spring be elongated by a displacement x due to force _F_1 and second spring be elongated by the same displacement x due to force _F_2, then the net force

$$[ F = -kx - k’x’ ]$$(10.46)

Equating equations (10.46) and (10.45), we get

$$[ F = -kx - k’x’ ]$$(10.47)

Generalizing, for n springs connected in parallel,

$$[ F = -kx - k’x’ ]$$(10.48)

a

If all spring constants are identical i.e., _k_1 = _k_2= … = kn = k then

$$k_1 = k_2 = \ldots = k_n = k \text{ then } \frac{k}{p} = n$$ (10.49)

This implies that the effective spring constant increases by a factor n. Hence, for the springs in parallel connection, the effective spring constant is greater than individual spring constant.

The spring constant is inversely proportional to the length of the spring

If the spring is cut into two pieces, one piece with length _l_1 and other with length _l_2, such that _l_1 = n_l_2, then spring constant of first length is

_k k n_1

1 =

+(n ) and spring constant of

second length is _k_2 = (n+1) k, where k is the original spring constant before cutting into pieces.

EXAMPLE 10.10

Consider two springs with force constants 1 N m−1 and 2 N m−1 connected in parallel. Calculate the effective spring constant (kp ) and comment on kp.

Solution

$$k_1 = 1, \text{Nm}^{-1},\quad k_2 = 2, \text{Nm}^{-1}$$

$$k_p = k_1 + k_2,, \text{Nm}^{-1}$$

$$k_p = 1 + 2 = 3, \text{Nm}^{-1}$$

$$k_p > k_1\quad \text{and}\quad k_p > k_2$$

Therefore, the effective spring constant is greater than both _k_1 and _k_2.

EXAMPLE 10.11

Calculate the equivalent spring constant for the following systems and also compute if all the spring constants are equal:

(a) (b)

Solution

a. Since _k_1 and _k_2 are parallel, $$k = k_1 + k_2 $$

Similarly, _k_3 and _k_4 are parallel, therefore, _k_d = _k_3 + _k_4

But _k_u and _k_d are in series,

therefore, $$k_{eq} = \frac{k_u}{k_d} / (k_u + k_d) $$

k keq u d

u d

= +

If all the spring constants are equal then, _k_1 = _k_2 = _k_3 = _k_4 = k

Which means, k_u = 2_k and k_d = 2_k

Hence, k k k

keq = = 4 4

2

b. Since _k_1 and _k_2 are parallel, _k_A = _k_1 + _k_2

Similarly, _k_4 and _k_5 are parallel, therefore, _k_B = _k_4 + _k_5

But _k_A, _k_3, _k_B, and _k_6 are in series,

therefore, 1 1 1 1 1

3 6_k k k k keq A B_

= + + +

If all the spring constants are equal then, _k_1 = _k_2 = _k_3 = _k_4 = _k_5 = _k_6 = k

which means, k_A = 2_k and k_B = 2_k

1 1 2

1 1 2

1 3 k k k k k keq

= + + + =

k k eq = 3

EXAMPLE 10.12

A mass m moves with a speed v on a horizontal smooth surface and collides with a nearly massless spring whose spring constant is k. If the mass stops after collision, compute the maximum compression of the spring.

Solution

When the mass collides with the spring, from the law of conservation of energy “the loss in kinetic energy of mass is gain in elastic potential energy by spring”. Let x be the distance of compression of spring, then the law of conservation of energy

$$[ \frac{1}{2}mv^2 = \frac{1}{2}kx^2 \Rightarrow x = \sqrt{\frac{m}{k}} ]$$

A pendulum is a mechanical system which exhibits periodic motion. It has a bob with mass m suspended by a long string (assumed to be massless and inextensible string) and the other end is fixed on a stand as shown in image 10.21 (a). At equilibrium, the pendulum does not oscillate and hangs vertically downward. Such a position is known as mean position or equilibrium position. When a pendulum is displaced through a small displacement from its equilibrium position and released, the bob of the pendulum executes to and fro motion. Let l be the length of the pendulum which is taken as the distance between the point of suspension and the centre of gravity of the bob. Two forces act on the bob of the pendulum at any displaced position, as shown in the image 10.21 (d),

(i) The gravitational force acting on the body (

) which acts vertically

downwards.

(ii) The tension in the string T 

which acts along the string to the point of suspension.

Resolving the gravitational force into its components:

a. Normal component: The component along the string but in opposition to the direction of tension, Fas = mg cosθ.

b. Tangential component: The component perpendicular to the string i.e., along tangential direction of arc of swing, Fps = mg sinθ.

Therefore, The normal component of the force is, along the string,

$$T - F_{\text{ds}} = \frac{m v^2}{l} \ \text{Here } v \text{ is speed of bob} \ T - mg \cos(\theta) = \frac{m v^2}{l} \quad (10.50) $$

From Newton’s 2nd law, 

_F ma_= Here, the net force on the L.H.S is T-Fas

In R.H.S, m _a_ is equivalent to the

centripetal force = mv^2/l

2

which makes the bob oscillate.

Note

From the image 10.21, we can observe that the tangential component Wps of the gravitational force always points towards the equilibrium position i.e., the direction in which it always points opposite to the direction of displacement of the bob from the mean position. Hence, in this case, the tangential force is nothing but the restoring force. Applying Newton’s second law along tangential direction, we have

$$m \frac{d^2s}{dt^2} + F_{ps} = 0 \Rightarrow m \frac{d^2s}{dt^2} = - F_{ps} \$$

$$m \frac{d^2s}{dt^2} = - mg\sin(\theta) \quad (10.51) $$

where, s is the position of bob which is measured along the arc. Expressing arc length in terms of angular displacement i.e., s = l θ (10.52) then its acceleration,

$$s = r\theta \quad (10.52) \

\frac{d^2s}{dt^2} = r\frac{d^2\theta}{dt^2} \quad (10.53) $$

Substituting equation (10.53) in equation (10.51), we get

$$\frac{d^2\theta}{dt^2} = -\frac{g}{l} \sin\theta \quad (10.53) \

\frac{d^2\theta}{dt^2} = -\frac{g}{l} \sin\theta \quad (10.54) $$

Because of the presence of sin θ in the above differential equation, it is a non-linear differential equation (Here, homogeneous second order). Assume “the small oscillation approximation”, sin θ ≈ θ, the above differential equation becomes linear differential equation.

$$\frac{d^2\theta}{dt^2} = -\frac{g}{l}\theta \quad (10.55) $$ This is the well known oscillatory differential equation. Therefore, the angular frequency of this oscillator (natural frequency of this system) is

$$\omega^2 = \frac{g}{l} \quad (10.56) \

\omega = \sqrt{\frac{g}{l}} \text{ rad s}^{-1} \quad (10.57) $$ The frequency of oscillations is

$$f = \frac{1}{2\pi} \sqrt{\frac{g}{l}} \quad \text{Hz} \quad (10.58) $$

and time period of oscillations is

$$T = 2\pi \sqrt{\frac{l}{g}} \text{ second} \quad (10.59) $$

Laws of simple pendulum The time period of a simple pendulum a. Depends on the following laws

(i) Law of length For a given value of acceleration due

to gravity, the time period of a simple pendulum is directly proportional to the square root of length of the pendulum.

_T l_µ (10.60)

(ii) Law of acceleration For a fixed length, the time period

of a simple pendulum is inversely proportional to square root of acceleration due to gravity.

T g

$$T \propto \frac{1}{\sqrt{g}} \quad (10.61) $$

b. Independent of the following factors (i) Mass of the bob

The time period of oscillation is independent of mass of the simple T l g

l = =2 2π π

pendulum. This is similar to free fall. Therefore, in a pendulum of fixed length, it does not matter whether an elephant swings or an ant swings. Both of them will swing with the same time period.

(ii) Amplitude of the oscillations For a pendulum with small angle

approximation (angular displacement is very small), the time period is independent of amplitude of the oscillation.

EXAMPLE 10.13

In simple pendulum experiment, we have used small angle approximation . Discuss the small angle approximation.

*This means that “for θ as large as 10 degrees, sin θ is nearly the same as θ when θ is expressed in radians”. As θ increases in value sinθ gradually becomes different from θ *

Pendulum length due to effect of temperature

Suppose the suspended wire is affected due to change in temperature. The rise in temperature affects length by

l = _l_o (1 + α ∆t) where l_o is the original length of the wire and l is final length of the wire when the temperature is raised. Let ∆_t is the change in temperature and α is the co-efficient of linear expansion.

Then, T l g

l g

l g

= = +

= +2 2 1 2 10 0π π α π α( t) ( t)∆ ∆

g

l g

+ = +

1 2 10 0α π α( t) ( t)∆ ∆

T T t T t= + ≈ +0

1 2

01 1 1 2

where ∆T is the change in time period due to the effect of temperature and T0 is the time period of the simple pendulum with original length _l_0.

Consider a U-shaped glass tube which consists of two open arms with uniform cross- sectional area A. Let us pour a non-viscous uniform incompressible liquid of density ρ in the U-shaped tube to a _h_eight h as shown in the image 10.22. If the liquid and tube are not disturbed then the liquid surface will be in equilibrium position O. It means the pressure as measured at any point on the liquid is the same and also at the surface on the arm (edge of the tube on either side), which balances with the atmospheric pressure. Due to this the level of liquid in each arm will be the same. By blowing air one can provide sufficient force in one arm, and the liquid gets disturbed from equilibrium position O, which means, the pressure at blown arm is higher than the other arm. This creates difference in pressure which will cause the liquid to oscillate for a very short duration of time about the mean or equilibrium position and finally comes to rest.

Time period of the oscillation is

T l g

= 2 2

$$T = 2\pi \sqrt{\frac{L}{2g}} \ \text{second} \quad (10.62) $$ Where l is the total length of liquid column in U- tube.

ENERGY IN SIMPLE HARMONIC MOTION

a. Expression for Potential Energy For the simple harmonic motion, the force and the displacement are related by Hooke’s law



$$\vec{F} = -k\vec{r}$$

Since force is a vector quantity, in three dimensions it has three components. Further, the force in the above equation is a conservative force field; such a force can be derived from a scalar function which has only one component. In one dimensional case

As we have discussed in unit 4 of volume I, the work done by the conservative force field is independent of path. The potential energy U can be calculated from the following expression.

$$\vec{F} = -k\vec{r}$$ =− (10.64)

Comparing (10.63) and (10.64), we get $$-\frac{dU}{dx} = -kx$$

$$dU = kx , dx $$

Dummy variable The integrating variable x’ (read x’ as “x prime”)is a dummy variable

$$\int_{0}^{y} t , dt = \int_{0}^{y} x , dx = \int_{0}^{y} p , dp = \frac{y^2}{2} $$

000 2 Notice that the integrating variables like t, x and p are dummy variables because, in this integration, whether we put t or x or p as variable for integration, we get the same answer. This work done by the force F during a small displacement dx stores as potential energy

From equation (10.22), we can substitute the value of force constant k = m ω2 in equation (10.65),

$$U(x) = \int_0^x k , dx = \frac{1}{2}k[x]^2_0 = \frac{1}{2}kx^2 \quad (10.65) $$

where ω is the natural frequency of the oscillating system. For the particle executing simple harmonic motion from equation (10.6), we get

$$U(x) = -\frac{1}{2} m \omega^2 x^2 \quad (10.66) $$

This variation of U is shown below.

Question to think over “If the potential energy is minimum then its second derivative is positive, why?”

b. Expression for Kinetic Energy Kinetic energy

$$\text{KE} = \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{dx}{dt}\right)^2 \quad (10.68) $$

Since the particle is executing simple harmonic motion, from equation (10.6)

x = A sin ω_t_ Therefore, velocity is

$$v_x = \frac{dx}{dt} = A\omega \cos(\omega t) \quad (10.69) \

v_x = A\omega\sqrt{1 - \left(\frac{x}{A}\right)^2} \

v_x = \omega\sqrt{A^2 - x^2} \quad (10.70) $$

Hence, $$v_x = \frac{dx}{dt} = A\omega \cos(\omega t) \quad (10.69) \

v_x = A\omega\sqrt{1 - \left(\frac{x}{A}\right)^2} \

v_x = \omega\sqrt{A^2 - x^2} \quad (10.70) $$

This variation with time is shown below.

c. Expression for Total Energy Total energy is the sum of kinetic energy and potential energy $$E = KE + U $$(10.73) $$E = \frac{1}{2}m\omega^2(A^2 - x^2) + \frac{1}{2}m\omega^2x^2 $$

Hence, cancelling _x_2 term,

$$E = \frac{1}{2} m \omega^2 A^2 = \text{constant} $$

Alternatively, from equation (10.67) and equation (10.72), we get the total energy as $$E = \frac{1}{2} m \omega^2 A^2 = \text{constant}$$

From trigonometry identity, (sin2 ω_t_ + cos2 ω_t)_ = 1

$$E = -\frac{1}{2} \mu \omega^2 A^2 = \text{constant} $$

which gives the law of conservation of total energy. This is depicted in image 10.26

Thus the amplitude of simple harmonic oscillator, can be expressed in terms of total energy.

$$A = \sqrt{\frac{2E}{m\omega^2}} = \sqrt{\frac{2E}{k}} \quad (10.75) $$

EXAMPLE 10.15

Write down the kinetic energy and total energy expressions in terms of linear momentum, For one-dimensional case.

Solution

Kinetic energy is KE mvx= 1 2

$$\text{Kinetic energy is } KE = \frac{1}{2}mv^2

\text{Multiply numerator and denominator by } m:

KE = \frac{\frac{1}{2}mv^2}{1} = \frac{(mv)^2}{2m} = \frac{(1v)^2}{2m} = \frac{p^2}{2m}

\text{where } p_x \text{ is the linear momentum of the particle experiencing simple harmonic motion.}

\text{Total energy can be written as sum of kinetic, from equation (10.73), and therefore from equation and potential energy, also from equation (10.75), we get:}

E=KE+U(x)=\frac{p_x^2}{2m} + \frac{1}{2}m\omega^2x^2 $$

Note

Conservation of energy Both the kinetic energy and potential energy areperiodic functions, and repeat their

values after a time period .

But total energy is constant for all the values of x or t. The kinetic energy and the potential energy for a simple harmonic motion are always positive. Note that kinetic energy cannot take negative value because it is proportional to the square of velocity. The measurement of any physical quantity must be a real number. Therefore, if kinetic energy is negative then the numerical value of velocity becomes an imaginary number, which is physically not acceptable. At equilibrium, it is purely kinetic energy and at extreme positions it is purely potential energy.

x = –A x = _x_0 = 0 x = A

Energy

$$U = \frac{1}{2}kx^2$$ $$For Potential Energy (U):

U = \frac{1}{2}kx^2$$ $$E_{total} = \frac{1}{2}kA^2$$ $$For Total Energy (E_total):

E_{total} = \frac{1}{2}kA^2$$

EXAMPLE 10.16

Compute the position of an oscillating particle when its kinetic energy and potential energy are equal.

Solution

Since the kinetic energy and potential energy of the oscillating particle are equal,

$$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x’^2$$

TYPES OF OSCILLATIONS:

Free oscillations

When the oscillator is allowed to oscillate by displacing its position from equilibrium position, it oscillates with a frequency which is equal to the natural frequency of the oscillator. Such an oscillation or vibration is known as free oscillation or free vibration. In this case, the amplitude, frequency and the energy of the vibrating object remains constant.

Examples:

(i) Vibration of a tuning fork.

(ii) Vibration in a stretched string.

(iii) Oscillation of a simple pendulum.

(iv) Oscillationsof a spring-mass system.

Damped oscillations

During the oscillation of a simple pendulum (in previous case), we have assumed that the amplitude of the oscillation is constant and also the total energy of the oscillator is constant. But in reality, in a medium, due to the presence of friction and air drag, the amplitude of oscillation decreases as time progresses. It implies that the oscillation is not sustained and the energy of the SHM decreases gradually indicating the loss of energy. The energy lost is absorbed by the surrounding medium. This type of

oscillatory motion is known as damped oscillation. In other words, if an oscillator moves in a resistive medium, its amplitude goes on decreasing and the energy of the oscillator is used to do work against the resistive medium. The motion of the oscillator is said to be damped and in this case, the resistive force (or damping force) is proportional to the velocity of the oscillator.

image 10.27 Damped harmonic oscillator – amplitude decreases as time increases.

Examples

(i) The oscillations of a pendulum (including air friction) or pendulum oscillating inside an oil filled container.

(ii) Electromagnetic oscillations in a tank circuit.

(iii) Oscillations in a dead beat and ballistic galvanometers.

Maintained oscillations

While playing in swing, the oscillations will stop after a few cycles, this is due to damping. To avoid damping we have to supply a push to sustain oscillations. By supplying energy from an external source, the amplitude of the oscillation can be made constant. Such vibrations are known as maintained vibrations.

Example: The vibration of a tuning fork getting energy from a battery or from external power supply.

Forced oscillations

Any oscillator driven by an external periodic agency to overcome the damping is known as forced oscillator or driven oscillator. In this type of vibration, the body executing vibration initially vibrates with its natural frequency and due to the presence of external periodic force, the body later vibrates with the frequency of the applied periodic force. Such vibrations are known as forced vibrations.

Example: Sound boards of stringed instruments.

Resonance

It is a special case of forced vibrations where the frequency of external periodic force (or driving force) matches with the natural frequency of the vibrating body (driven). As a result the oscillating body begins to vibrate such that its amplitude increases at each step and ultimately it has a large amplitude. Such a phenomenon is known as resonance and the corresponding vibrations are known as resonance vibrations.

Example

The breaking of glass due to sound

Note The concept of resonance is used in Tuning of station (or channel) in a radio (or Television) circuits.

Extra: Pendulum in a lift:

(i) Lift moving upwards with acceleration

Effective acceleration due to gravity is

$$g_{eff} = g + a$$

$$T = 2\pi \sqrt{\frac{l}{g_{eff}}} = 2\pi \sqrt{\frac{l}{g + a}}$$

Since the time period is inversely relat period will decrease when lift moves u

(ii) Lift moving downwards with accelerat Effective acceleration due to gravity is

Then time period is $$T = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{l}{g - a}}$$

Since the time period is inversely relat period will increase when lift moves do

(iii) Lift falls with acceleration a > g: The effective acceleration is$$g_{\text{eff}} = a - g$$ $$T = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{l}{a - g}}$$

in this case, the pendulum will turn upsi

(iv) Lift falls with acceleration a = g:

The effective acceleration is geff = g - g = Then time period is T → ∞ which means arrested.

(v) If the simple pendulum is kept in a car w The effective acceleration is $$g_{\text{eff}} = \sqrt{g^2 + a^2}$$

$$T = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{l}{\sqrt{g^2 + a^2}}}$$

I. Multiple Choice Questions

1. In a simple harmonic oscillation, the acceleration against displacement for one complete oscillation will be

(model NSEP 2000-01) a) an ellipse b) a circle c) a parabola d) a straight line

2. A particle executing SHM crosses points A and B with the same velocity. Having taken 3 s in passing from A to B, it returns to B after another 3 s. The time period is

a) 15 s b) 6 s c) 12 s d) 9 s

3. The length of a second’s pendulum on the surface of the Earth is 0.9 m. The length of the same pendulum on surface of planet X such that the acceleration of the planet X is n times greater than the Earth is

a) 0.9_n_ b) 0 9 /n m

c) 0.9 n^2 m^2 d) 0.9/n^2

.

n 4. A simple pendulum is suspended from

the roof of a school bus which moves in a horizontal direction with an acceleration a, then the time period is

a)$$ ( T \propto \frac{1}{g^2 + a^2} )$$

b)$$( T \propto \sqrt{\frac{1}{g^2 + a^2}} )$$

c)$$( T \propto \sqrt{g^2 + a^2} )$$

d)$$( T \propto (g^2 + a^2) $$

5. Two bodies A and B whose masses are in the ratio 1:2 are suspended from two separate massless springs of force constants kA and kB respectively. If the two bodies oscillate vertically such that

EVALUATION

their maximum velocities are in the ratio 1:2, the ratio of the amplitude A to that of B is

a)$$( \sqrt{\frac{k_B}{2k_A}} )$$

b) $$( \sqrt{\frac{k_B}{8k_A}} )$$

c) $$( \sqrt{\frac{2k_B}{k_A}} )$$

d) $$( \sqrt{\frac{8k_B}{k_A}} )$$

6. A spring is connected to a mass m suspended from it and its time period for vertical oscillation is T. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillation is

a) $$( T’ = \sqrt{2}T )$$

b) $$( T’ = \frac{T}{\sqrt{2}} )$$

c) $$( T’ = \sqrt{2T} )$$

d) $$( T’ = \frac{\sqrt{T}}{2} )$$

7. The displacement of a simple harmonic motion is given by y(t) = A sin (ω_t_ + ϕ) where A is amplitude of the oscillation, ω is the angular frequency and ϕ is the phase. Let the amplitude of the oscillation be 8 cm and the time period of the oscillation is 24 s. If the displacement at initial time (t = 0 s) is 4 cm, then the displacement at t = 6 s is

(a) 8 cm

(b) 4 cm

(c) 3 cm4

(d) 3 cm8 8. A simple pendulum has a time period _T_1. When its point of suspension is moved vertically upwards according as y = _k t_2, where y is vertical distance covered and k = 1 _ms_−2, its time period becomes T2. Then, T

T 1 2

2 2

is (g = 10 m s−2) (IIT 2005) a) ( \frac{5}{6} )$$

b)$$ ( \frac{11}{10} )$$

c) $$( \frac{6}{5} )$$

d) $$( -\frac{5}{4} )$$

9. An ideal spring of spring constant k, is suspended from the ceiling of a room and a block of mass M is fastened to its lower end. If the block is released when the spring is un-stretched, then the maximum extension in the spring is (IIT 2002)

a) 4 Mg k

b) Mg k

c) 2 Mg k

d) _Mg k_2

10. A pendulum is hung in a very high building oscillates to and fro motion freely like a simple harmonic oscillator. If the acceleration of the bob is 16 _ms_−2 at a distance of 4 m from the mean position, then the time period is (NEET 2018 model) a) 2 s b) 1 s c) 2πs (d) πs

11. A hollow sphere is filled with water. It is hung by a long thread. As the water flows out of a hole at the bottom, the period of oscillation will

a) first increase and then decrease b) first decrease and then increase c) increase continuously d) decrease continuously

12. The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are (AIPMT 2012)

a) kg m _s_−1 b) kg m _s_−2

c) kg _s_−1 (d) kg s

13. Let the total energy of a particle executing simple harmonic motion with angular frequency is 1 rad s–1 is 0.256 J. If the displacement of the particle at time t =

π 2

s is 2 cm8 then the amplitude of

motion is

a) 8 cm b) 16 cm

c) 32 cm d) 64 cm

14. A particle executes simple harmonic motion and displacement y at time _t_0, 2_t_0 and 3_t_0 are A, B and C, respectively. Then the value of ∙A + C

2_B_ ∙ is $$a) ( \cos(\omega t_0) )$$

$$b) ( \cos(2\omega t_0) )$$

$$c) ( \cos(3\omega t_0) )$$

d) 1 15. A mass of 3 kg is attached at the end of a spring moves with simple harmonic motion on a horizontal frictionless table with time period 2π and with amplitude of 2m, then the maximum fore exerted on the spring is

(a) 1.5 N (b) 3 N

(c) 6 N (d) 12 N

Answers:

1. d 2) c 3) a 4) b 5) b 6) b 7) d 8) c 9) c 10) d 11) a 12) c 13) b 14) a 15) c II. Short Answers Questions

1. What is meant by periodic and non- periodic motion?. Give any two examples, for each motion.

2. What is meant by force constant of a spring?.

3. Define time period of simple harmonic motion.

4. Define frequency of simple harmonic motion.

5. What is an epoch?. 6. Write short notes on two springs

connected in series. 7. Write short notes on two springs

connected in parallel. 8. Write down the time period of simple

pendulum. 9. State the laws of simple pendulum?. 10. Write down the equation of time period

for linear harmonic oscillator. 11. What is meant by free oscillation?. 12. Explain damped oscillation. Give an

example. 13. Define forced oscillation. Give an

example. 14. What is meant by maintained

oscillation?. Give an example. 15. Explain resonance. Give an example.

III. Long Answers Questions

1. What is meant by simple harmonic oscillation?. Give examples and explain why every simple harmonic motion is a periodic motion whereas the converse need not be true.

2. Describe Simple Harmonic Motion as a projection of uniform circular motion.

3. What is meant by angular harmonic oscillation?. Compute the time period of angular harmonic oscillation.

4. Write down the difference between simple harmonic motion and angular simple harmonic motion.

5. Discuss the simple pendulum in detail. 6. Explain the horizontal oscillations of a

spring. 7. Describe the vertical oscillations of a

spring. 8. Write short notes on the oscillations of

liquid column in U-tube. 9. Discuss in detail the energy in simple

harmonic motion. 10. Explain in detail the four different

types of oscillations.

IV. Exercises 1. Given an one dimensional system with

total energy E = p_2_x 2_m_

+ V(x) = constant,

where px is the x component of the linear momentum and V(x) is the potential energy of the system. Show that total time derivative of energy gives us force

Fx = – d

dx V(x). Verify Hooke’s law by

choosing potential energy V(x) = 1 2

_kx_2.

2. Consider a simple pendulum of length l = 0.9 m which is properly placed on a trolley rolling down on a inclined plane which is at θ = 45° with the horizontal. Assuming that the inclined plane is frictionless, calculate the time period of oscillation of the simple pendulum.

Answer: 0.86 s 3. A piece of wood of mass m is floating

erect in a liquid whose density is ρ. If it is slightly pressed down and released, then executes simple harmonic motion. Show that its time period of oscillation

is T m Ag

=2π ρ

4. Consider two simple harmonic motion along x and y-axis having same frequencies but different amplitudes as x = A sin (ω_t_ + φ) (along x axis) and y = B sin ωt (along y axis). Then show that

$$[ \frac{x2}{A2} + \frac{y2}{B2} - \frac{2xy}{AB}\cos\phi = \sin^2\phi ]$$

and also discuss the special cases when

a. φ = 0 b. φ = π c. ϕ π =

2

d. ϕ π =

2 and A = B (e) ϕ π

= 4

Note: when a particle is subjected to two simple harmonic motion at right angle to each other the particle may move along different paths. Such paths are called Lissajous images. Answer :

a. y B A

x= , equation is a straight line

passing through origin with positive slope.

b. y B A

x=− equation is a straight line passing through origin with negative slope.

c. x A

y B

2

2

2

2 1+ = , equation is an ellipse whose

center is origin. d. _x_2+_y_2 = _A_2, equation is a circle whose

center is origin .

a) $$( y = \frac{A}{x} )$$

b) $$( y = -\frac{B}{x} )$$

c)$$ ( \frac{x2}{A2} + \frac{y2}{B2} = 1 )$$

d) $$( x^2 + y^2 = A^2 )$$

e) $$( \frac{x2}{A2} - \frac{y2}{B2} = 1 - 2xy\sqrt{\frac{1}{AB}} - \frac{1}{A^{3/4}}\sqrt[5]{\frac{1}{AB}} + 2^{3/4} )$$

ellipse (oblique ellipse which means tilted ellipse)

5. Show that for a particle executing simple harmonic motion

a. the average value of kinetic energy is equal to the average value of potential energy.

b. average potential energy = average

kinetic energy = 1 2

(total energy)

Hint : average kinetic energy = <kinetic

energy> = 1 0_T_

Kineticenergy dt T ( )∫

and

average Potential energy = <Potential

energy> = 1 0_T_

Potential energy dt T ( )∫

6. Compute the time period for the following system if the block of mass m is slightly displaced vertically down from its equilibrium position and then released. Assume that the pulley is light and smooth, strings and springs are light.

k2

k1

m

k

m

k

m

k2

k1

m

k

m

k

m Hint and answer:

Case(a)

Pulley is fixed rigidly here. When the mass displace by y and the spring will also stretch by y. Therefore, F = T = ky

T m k

= 2π

BOOKS FOR REFERENCE

1. Vibrations and Waves – A. P. French, CBS p

2. Concepts of Physics – H. C. Verma, Volume

3. Fundamentals of Physics – Halliday, Resnick

4. Physics for Scientist and Engineers with Mo Coole Publishers, Eighth Edition.

Case(b)

Mass displace by y, pulley also displaces by y. T = 4_ky_.

T m k

= 2 4

π

ublisher and Distributors Pvt. Ltd.

1 and Volume 2, Bharati Bhawan Publisher.

and Walker, Wiley Publishers, 10th edition.

dern Physics – Serway and Jewett, Brook/ Oscilla

Through this activity you will be able to learn about the resonance.

STEPS:

• Use the URL or scan the QR code to open ‘Ph

button.

• In the activity window a diagram of resonato slider on ‘sim speed’ given below to see the re

• Move the slider to change ‘Number of Resona side window and see the ‘frequency’.

• Select the ‘On’, ‘Off ’ button on ‘Gravity’ to see

URL: https://phet.colorado.edu/en/simulation/legacy/resona

* Pictures are indicative only. * If browser requires, allow Flash Player or Java Scr

ICT CORNER

Step1

Step3

tions

ET’ simulation on ‘Resonance’. Click the play

r is given. Click the play icon and move the sonance.

tors’, ‘Mass’ and ‘Spring constant’ on the right

the different resonance.

nce

ipt to load the page.

Step4

Step2