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Oscillations Class 11 Physics Notes

Introduction

In physics, we study many types of motion. Some, like a ball thrown in the air, are non-repetitive. Others, like the orbit of a planet, repeat over a fixed time interval. This repeating motion is called periodic motion.

A special kind of periodic motion is oscillatory motion, where an object moves back and forth about a central, or "mean," position. Think of a child on a swing, the pendulum of a clock, or a guitar string when it's plucked. These are all examples of oscillations.

The study of oscillations is fundamental because it helps us understand a wide range of physical phenomena, from the vibrations of atoms in a solid to the propagation of sound and light waves. To describe these motions, we use key concepts like:

Period: The time for one complete oscillation.
Frequency: How many oscillations occur per second.
Displacement: The object's position relative to its mean position.
Amplitude: The maximum displacement.
Phase: The state of the object's motion at a specific time.

Periodic and Oscillatory Motions

A motion that repeats itself at regular intervals of time is called periodic motion. The smallest interval of time after which the motion repeats is called its period (T).

Often, a body in periodic motion has an equilibrium position, where no net force acts on it. If you move the body slightly from this position, a restoring force tries to pull it back. This force causes the body to move back and forth, or oscillate, around the equilibrium point. This is called oscillatory motion or vibration.

Note

Every oscillatory motion is periodic, but not every periodic motion is oscillatory. For example, a planet's orbit is periodic, but it doesn't move back and forth about a mean position, so it's not oscillatory.

Period and frequency

Period (T) is the time taken to complete one full oscillation. Its SI unit is the second (s).

Frequency (v) is the number of oscillations that occur per unit time. It is the reciprocal of the period. $v = \frac{1}{T}$ The SI unit of frequency is $\text{s}^{-1}$ , which is named the hertz (Hz) in honor of Heinrich Hertz, the discoverer of radio waves. $1 \text{ hertz} = 1 \text{ Hz} = 1 \text{ oscillation per second} = 1 \text{ s}^{-1}$

Example

On an average, a human heart is found to beat 75 times in a minute. Calculate its frequency and period.

Given

Number of beats = 75
Time = 1 minute = 60 s

To Find

Frequency (v)
Period (T)

Formula

$v = \frac{\text{Number of repetitions}}{\text{Total time}}$ $T = \frac{1}{v}$

Solution

First, calculate the frequency in beats per second (Hz).

$v = \frac{75}{60 \text{ s}} = 1.25 \text{ s}^{-1} = 1.25 \text{ Hz}$

Next, calculate the period using the frequency.

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

Final Answer The frequency of the heart is $1.25 \text{ Hz}$ , and the time period is $0.8 \text{ s}$ .

Displacement

In the context of oscillations, displacement refers to the change with time of any physical property. It's not just about position. It could be the angle of a pendulum, the voltage in an AC circuit, or the pressure in a sound wave.

Displacement in periodic motion can be represented by a periodic function of time. The simplest periodic functions are sine and cosine. For example, a function like $f(t) = A \cos \omega t$ is periodic. Its period, T, is given by: $T = \frac{2\pi}{\omega}$ where $\omega$ is the angular frequency.

According to Fourier's theorem, any periodic function can be expressed as a combination of sine and cosine functions.

Example

Which of the following functions of time represent (a) periodic and (b) non-periodic motion? Give the period for each case of periodic motion [

\omega

is any positive constant]. (i)

\sin \omega t + \cos \omega t

(ii)

\sin \omega t + \cos 2 \omega t + \sin 4 \omega t

(iii)

e^{-\omega t}

(iv)

\log(\omega t)

Solution

(i) $\sin \omega t + \cos \omega t$

This function is periodic. It can be rewritten as $\sqrt{2} \sin(\omega t + \pi/4)$ . Since the value of the sine function repeats every $2\pi$ radians, the motion repeats when $\omega t$ increases by $2\pi$ . The time period $T$ is the time it takes for this to happen: $\omega T = 2\pi$ , so $T = 2\pi/\omega$ .

(ii) $\sin \omega t + \cos 2 \omega t + \sin 4 \omega t$

This is also a periodic motion. Each term is a periodic function with a different period:

$\sin \omega t$ has a period $T_0 = 2\pi/\omega$ .
$\cos 2\omega t$ has a period $T_0/2 = \pi/\omega$ .
$\sin 4\omega t$ has a period $T_0/4 = \pi/(2\omega)$ .

The period of the combined function is the least common multiple of the individual periods. Since the first term's period ( $T_0$ ) is a multiple of the other two, the entire function repeats with a period of $T_0 = 2\pi/\omega$ .

(iii) $e^{-\omega t}$

This function is not periodic. It continuously decreases as time increases and approaches zero, but it never repeats its value.

(iv) $\log(\omega t)$

This function is not periodic. It increases continuously with time and never repeats its value.

Simple Harmonic Motion

An object oscillating back and forth about an origin is said to be in Simple Harmonic Motion (SHM) if its displacement $x$ from the origin varies with time as a sinusoidal function: $x(t) = A \cos(\omega t + \phi)$ where $A$ , $\omega$ , and $\phi$ are constants.

Let's break down the terms in this equation:

$x(t)$ is the displacement from the equilibrium position at time $t$ .
Amplitude (A) is the magnitude of the maximum displacement of the particle. The motion is confined between $-A$ and $+A$ .
Angular Frequency (ω) is related to how fast the oscillations occur. It is related to the period $T$ and frequency $v$ by: $\omega = \frac{2\pi}{T} = 2\pi v$ Its SI unit is radians per second ( $\text{rad s}^{-1}$ ).
Phase (ωt + φ) is the argument of the cosine function. It determines the state of the motion (position and velocity) at any time $t$ .
Phase Constant (φ), also called the phase angle, is the value of the phase at $t=0$ . It tells us the starting position of the particle.

Example

Which of the following functions of time represent (a) simple harmonic motion and (b) periodic but not simple harmonic? Give the period for each case. (1)

\sin \omega t - \cos \omega t

(2)

\sin^2 \omega t

Solution

(a) $\sin \omega t - \cos \omega t$

This function can be rewritten in the standard form of SHM: $\sin \omega t - \cos \omega t = \sqrt{2} \sin(\omega t - \pi/4)$ This represents a simple harmonic motion with a period $T = 2\pi/\omega$ and a phase constant of $-\pi/4$ .

(b) $\sin^2 \omega t$

Using the trigonometric identity $\cos 2\theta = 1 - 2\sin^2\theta$ , we can rewrite the function as: $\sin^2 \omega t = \frac{1}{2} - \frac{1}{2} \cos 2\omega t$ This function is periodic. The term $\cos 2\omega t$ has an angular frequency of $2\omega$ , so the period is $T = 2\pi / (2\omega) = \pi/\omega$ . However, this is not simple harmonic motion in the standard sense because the equilibrium point is shifted to $1/2$ instead of zero. It is a harmonic motion about a different equilibrium position.

Simple Harmonic Motion and Uniform Circular Motion

There's a fascinating connection between SHM and uniform circular motion. The projection of a particle moving uniformly on a circle onto any diameter of that circle executes SHM.

Imagine a particle P moving in a circle of radius $A$ with a constant angular speed $\omega$ .

The circle is called the reference circle.
The particle P is called the reference particle.

At any time $t$ , the angle the position vector OP makes with the x-axis is $(\omega t + \phi)$ , where $\phi$ is the initial angle at $t=0$ .

The projection of the particle's position onto the x-axis, let's call it P', has a position given by: $x(t) = A \cos(\omega t + \phi)$ This is the defining equation for SHM. This shows that as P moves in a circle, its "shadow" or projection on the x-axis moves back and forth in simple harmonic motion.

Similarly, the projection onto the y-axis is: $y(t) = A \sin(\omega t + \phi)$ This is also an SHM, with the same amplitude and frequency, but with a phase difference of $\pi/2$ from the x-projection.

Note

Although SHM and uniform circular motion are related, the forces involved are very different. The force in SHM is a restoring force (

F = -kx

), while the force in uniform circular motion is a constant-magnitude centripetal force.

Example

The figure given below depicts two circular motions. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated in the figures. Obtain the simple harmonic motions of the x-projection of the radius vector of the rotating particle P in each case.

Given

(a)

Radius = $A$
Period, $T = 4$ s
Initial angle at $t=0$ , $\phi = 45^\circ = \pi/4$ rad
Sense of revolution: anticlockwise

(b)

Radius = $B$
Period, $T = 30$ s
Initial angle at $t=0$ , $\phi = 90^\circ = \pi/2$ rad
Sense of revolution: clockwise

To Find

The equation for the x-projection of the particle's motion, $x(t)$ , in each case.

Formula

The general equation for the x-projection is $x(t) = A \cos(\theta)$ , where $\theta$ is the angle with the positive x-axis at time $t$ . Angular speed, $\omega = \frac{2\pi}{T}$ .

Solution

(a) The particle starts at an angle of $\pi/4$ . It moves anticlockwise, so the angle increases with time. The angle at time $t$ is: $\theta(t) = \omega t + \phi = \frac{2\pi}{T}t + \frac{\pi}{4}$ Substituting the value of $T = 4$ s: $\theta(t) = \frac{2\pi}{4}t + \frac{\pi}{4} = \frac{\pi}{2}t + \frac{\pi}{4}$ The x-projection is therefore: $x(t) = A \cos\left(\frac{\pi}{2}t + \frac{\pi}{4}\right)$ This is an SHM with amplitude $A$ , period 4 s, and initial phase $\pi/4$ .

(b) The particle starts at an angle of $\pi/2$ . It moves clockwise, so the angle decreases with time. The angle at time $t$ is: $\theta(t) = \phi - \omega t = \frac{\pi}{2} - \frac{2\pi}{T}t$ Substituting the value of $T = 30$ s: $\theta(t) = \frac{\pi}{2} - \frac{2\pi}{30}t = \frac{\pi}{2} - \frac{\pi}{15}t$ The x-projection is therefore: $x(t) = B \cos\left(\frac{\pi}{2} - \frac{\pi}{15}t\right)$ Using the identity $\cos(\pi/2 - \theta) = \sin(\theta)$ , this can be written as: $x(t) = B \sin\left(\frac{\pi}{15}t\right)$ This represents an SHM with amplitude $B$ and period 30 s. To write it in the standard cosine form, we use $\sin(\theta) = \cos(\theta - \pi/2)$ , which gives an initial phase of $-\pi/2$ .

Velocity and Acceleration in Simple Harmonic Motion

We can find the velocity and acceleration of a particle in SHM by differentiating the displacement equation $x(t) = A \cos(\omega t + \phi)$ with respect to time.

Velocity

Differentiating displacement gives velocity: $v(t) = \frac{dx}{dt} = \frac{d}{dt}[A \cos(\omega t + \phi)]$ $v(t) = -\omega A \sin(\omega t + \phi)$ Key points about velocity in SHM:

The maximum speed (velocity amplitude) is $v_m = \omega A$ .
Velocity is maximum at the equilibrium position ( $x=0$ ) and zero at the extreme positions ( $x = \pm A$ ).
The velocity function is out of phase with the displacement by $\pi/2$ radians (one-quarter of a cycle).

Acceleration

Differentiating velocity gives acceleration: $a(t) = \frac{dv}{dt} = \frac{d}{dt}[-\omega A \sin(\omega t + \phi)]$ $a(t) = -\omega^2 A \cos(\omega t + \phi)$ Since $x(t) = A \cos(\omega t + \phi)$ , we can write a very important relationship: $a(t) = -\omega^2 x(t)$ Key points about acceleration in SHM:

The acceleration is directly proportional to the displacement.
The negative sign shows that the acceleration is always directed opposite to the displacement, meaning it always points towards the equilibrium position.
The maximum acceleration (acceleration amplitude) is $a_m = \omega^2 A$ .
Acceleration is maximum at the extreme positions ( $x = \pm A$ ) and zero at the equilibrium position ( $x=0$ ).
The acceleration function is out of phase with the displacement by $\pi$ radians (half a cycle).

Example

A body oscillates with SHM according to the equation (in SI units),

x = 5 \cos(2\pi t + \pi/4)

. At

t = 1.5

s, calculate the (a) displacement, (b) speed and (c) acceleration of the body.

Given

Displacement equation: $x = 5 \cos(2\pi t + \pi/4)$
Time, $t = 1.5$ s

From the equation, we can identify:

Amplitude, $A = 5$ m
Angular frequency, $\omega = 2\pi \text{ rad s}^{-1}$
Phase constant, $\phi = \pi/4$ rad

To Find

(a) Displacement, $x$ (b) Speed, $v$ (c) Acceleration, $a$

Formula

$x(t) = 5 \cos(2\pi t + \pi/4)$ $v(t) = -\omega A \sin(\omega t + \phi) = -(2\pi)(5) \sin(2\pi t + \pi/4)$ $a(t) = -\omega^2 x(t) = -(2\pi)^2 x(t)$

Solution

First, let's evaluate the phase at $t = 1.5$ s: $\text{Phase} = 2\pi(1.5) + \pi/4 = 3\pi + \pi/4$ Since $\cos(3\pi + \theta) = -\cos(\theta)$ and $\sin(3\pi + \theta) = -\sin(\theta)$ , we have: $\cos(3\pi + \pi/4) = -\cos(\pi/4) = -1/\sqrt{2} \approx -0.707$ $\sin(3\pi + \pi/4) = -\sin(\pi/4) = -1/\sqrt{2} \approx -0.707$

(a) Displacement at $t=1.5$ s $x = 5 \cos(3\pi + \pi/4) = 5 \times (-0.707) \text{ m}$ $x = -3.535 \text{ m}$ Answer for part (a) = $-3.535$ m

(b) Speed at $t=1.5$ s The velocity is $v(t) = -10\pi \sin(2\pi t + \pi/4)$ . $v = -10\pi \sin(3\pi + \pi/4) = -10\pi \times (-0.707) \text{ m s}^{-1}$ $v \approx 22.21 \text{ m s}^{-1}$ The speed is the magnitude of the velocity. Answer for part (b) = $22 \text{ m s}^{-1}$ (rounded)

(c) Acceleration at $t=1.5$ s $a = -(2\pi)^2 x = -4\pi^2 \times (-3.535) \text{ m s}^{-2}$ $a \approx 139.6 \text{ m s}^{-2}$ Answer for part (c) = $140 \text{ m s}^{-2}$ (rounded)

Force Law for Simple Harmonic Motion

According to Newton's second law, force is mass times acceleration ( $F=ma$ ). For a particle of mass $m$ in SHM, we have: $F(t) = m \times a(t) = m \times [-\omega^2 x(t)]$ $F(t) = -m\omega^2 x(t)$ This is the force law for SHM. We can define a constant $k = m\omega^2$ , called the force constant or spring constant. The equation then becomes: $F(t) = -k x(t)$ This is also known as Hooke's Law. It states that for SHM, the restoring force is directly proportional to the displacement and is always directed towards the mean (equilibrium) position.

From $k = m\omega^2$ , we can express the angular frequency and period in terms of the system's physical properties (mass and force constant): $\omega = \sqrt{\frac{k}{m}}$ $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{k}}$ A particle oscillating under such a force is called a linear harmonic oscillator.

Example

Two identical springs of spring constant

k

are attached to a block of mass

m

and to fixed supports as shown in the figure. Show that when the mass is displaced from its equilibrium position on either side, it executes a simple harmonic motion. Find the period of oscillations.

Given

Mass of the block = $m$
Spring constant of each spring = $k$

To Find

Show the motion is SHM.
Find the period of oscillation, $T$ .

Solution

Let the equilibrium position be $x=0$ . Suppose we displace the mass by a small distance $x$ to the right.

The left spring is stretched by $x$ . It exerts a restoring force to the left: $F_1 = -kx$ .
The right spring is compressed by $x$ . It exerts a restoring force by pushing to the left: $F_2 = -kx$ .

The net force on the mass is the sum of these two forces: $F_{net} = F_1 + F_2 = -kx - kx = -2kx$ The net force is $F_{net} = -(2k)x$ .

This equation is in the form $F = -k_{eff}x$ , where the effective spring constant is $k_{eff} = 2k$ . Since the net restoring force is directly proportional to the displacement and directed towards the mean position, the motion is simple harmonic.

The period of oscillation is given by the formula $T = 2\pi\sqrt{m/k_{eff}}$ . $T = 2\pi\sqrt{\frac{m}{2k}}$

Final Answer The motion is simple harmonic with a period of $T = 2\pi\sqrt{\frac{m}{2k}}$ .

Energy in Simple Harmonic Motion

A particle in SHM possesses both kinetic and potential energy.

Kinetic Energy (K)

The kinetic energy is given by $K = \frac{1}{2}mv^2$ . Substituting the expression for velocity, $v(t) = -\omega A \sin(\omega t + \phi)$ : $K = \frac{1}{2}m [-\omega A \sin(\omega t + \phi)]^2$ $K = \frac{1}{2}m\omega^2 A^2 \sin^2(\omega t + \phi)$ Since $k = m\omega^2$ , this can also be written as: $K = \frac{1}{2}kA^2 \sin^2(\omega t + \phi)$ Kinetic energy is maximum ( $\frac{1}{2}kA^2$ ) at the mean position ( $x=0$ ) and zero at the extreme positions ( $x=\pm A$ ).

Potential Energy (U)

The restoring force $F = -kx$ is a conservative force. The potential energy associated with this force is: $U = \frac{1}{2}kx^2$ Substituting the expression for displacement, $x(t) = A \cos(\omega t + \phi)$ : $U = \frac{1}{2}k [A \cos(\omega t + \phi)]^2$ $U = \frac{1}{2}kA^2 \cos^2(\omega t + \phi)$ Potential energy is maximum ( $\frac{1}{2}kA^2$ ) at the extreme positions ( $x=\pm A$ ) and zero at the mean position ( $x=0$ ).

Total Energy (E)

The total mechanical energy is the sum of kinetic and potential energies: $E = K + U = \frac{1}{2}kA^2 \sin^2(\omega t + \phi) + \frac{1}{2}kA^2 \cos^2(\omega t + \phi)$ Using the identity $\sin^2\theta + \cos^2\theta = 1$ : $E = \frac{1}{2}kA^2 [ \sin^2(\omega t + \phi) + \cos^2(\omega t + \phi) ]$ $E = \frac{1}{2}kA^2$ The total mechanical energy of a simple harmonic oscillator is constant and does not depend on time or displacement. It is proportional to the square of the amplitude.

Example

A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of

50 \text{ N m}^{-1}

. The block is pulled to a distance

x = 10

cm from its equilibrium position at

x=0

on a frictionless surface from rest at

t=0

. Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position.

Given

Mass, $m = 1$ kg
Spring constant, $k = 50 \text{ N m}^{-1}$
Amplitude (maximum displacement), $A = 10 \text{ cm} = 0.1$ m
Position for calculation, $x = 5 \text{ cm} = 0.05$ m

To Find

Kinetic energy (K), Potential energy (U), and Total energy (E) at $x = 0.05$ m.

Formula

$U = \frac{1}{2}kx^2$ $E = \frac{1}{2}kA^2$ $K = E - U$

Solution

First, let's calculate the total energy of the system, which remains constant throughout the motion. $E = \frac{1}{2}kA^2 = \frac{1}{2} \times (50 \text{ N m}^{-1}) \times (0.1 \text{ m})^2$ $E = \frac{1}{2} \times 50 \times 0.01 = 0.25 \text{ J}$ The total energy is $0.25$ J.

Now, calculate the potential energy when the block is at $x = 0.05$ m. $U = \frac{1}{2}kx^2 = \frac{1}{2} \times (50 \text{ N m}^{-1}) \times (0.05 \text{ m})^2$ $U = \frac{1}{2} \times 50 \times 0.0025 = 0.0625 \text{ J}$ The potential energy at this position is $0.0625$ J.

Finally, calculate the kinetic energy at this position. The kinetic energy is the difference between the total energy and the potential energy. $K = E - U = 0.25 \text{ J} - 0.0625 \text{ J} = 0.1875 \text{ J}$ The kinetic energy at this position is approximately $0.19$ J.

Final Answer

Potential Energy (at $x=5$ cm) = $0.0625$ J
Kinetic Energy (at $x=5$ cm) = $0.1875$ J (or $0.19$ J)
Total Energy = $0.25$ J

The Simple Pendulum

A simple pendulum consists of a small, heavy bob of mass $m$ attached to a long, inextensible, massless string of length $L$ , fixed at the other end. When displaced slightly from its vertical equilibrium position, it oscillates back and forth.

Let's analyze the forces acting on the bob when it is displaced by an angle $\theta$ from the vertical.

The force of gravity, $mg$ , acts vertically downwards.
The tension, $T$ , acts along the string.

We can resolve the gravitational force into two components:

$mg \cos\theta$ : along the string, balancing the tension.
$mg \sin\theta$ : perpendicular to the string, acting as the restoring force that brings the bob back to the equilibrium position.

The restoring torque ( $\tau$ ) about the support point is caused by the tangential force component: $\tau = -L(mg \sin\theta)$ The negative sign indicates that the torque acts to reduce the angular displacement $\theta$ .

According to Newton's law for rotation, $\tau = I\alpha$ , where $I$ is the moment of inertia and $\alpha$ is the angular acceleration. $I\alpha = -mgL \sin\theta$ $\alpha = -\frac{mgL}{I} \sin\theta$ For a simple pendulum, the moment of inertia of the bob about the support is $I = mL^2$ . $\alpha = -\frac{mgL}{mL^2} \sin\theta = -\frac{g}{L} \sin\theta$ This equation is not in the standard form for SHM because of the $\sin\theta$ term. However, if the angle $\theta$ is small (typically less than about 15 degrees), we can use the small-angle approximation: $\sin\theta \approx \theta$ (where $\theta$ is in radians).

With this approximation, the equation becomes: $\alpha = -\left(\frac{g}{L}\right)\theta$ This equation is mathematically identical to the acceleration equation for linear SHM ( $a = -\omega^2 x$ ), with angular acceleration $\alpha$ replacing linear acceleration $a$ , and angular displacement $\theta$ replacing linear displacement $x$ .

Therefore, for small oscillations, the motion of a simple pendulum is simple harmonic. By comparing the two equations, we see that: $\omega^2 = \frac{g}{L} \quad \implies \quad \omega = \sqrt{\frac{g}{L}}$ The period of oscillation for a simple pendulum is: $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L}{g}}$

Note

The period of a simple pendulum depends only on its length (

L

) and the acceleration due to gravity (

g

). It does not depend on the mass of the bob or the amplitude of the oscillation (as long as the amplitude is small).

Example

What is the length of a simple pendulum, which ticks seconds?

Given

A pendulum that "ticks seconds" completes one half of an oscillation in one second. Therefore, its full period is the time for a tick and a tock.

Period, $T = 2$ s
Acceleration due to gravity, $g = 9.8 \text{ m s}^{-2}$

To Find

The length of the pendulum, $L$ .

Formula

$T = 2\pi\sqrt{\frac{L}{g}}$

Solution

We need to rearrange the formula to solve for $L$ . First, square both sides: $T^2 = (2\pi)^2 \frac{L}{g} = 4\pi^2 \frac{L}{g}$ Now, isolate $L$ : $L = \frac{gT^2}{4\pi^2}$ Substitute the given values: $L = \frac{(9.8 \text{ m s}^{-2})(2 \text{ s})^2}{4\pi^2} = \frac{9.8 \times 4}{4\pi^2} = \frac{9.8}{\pi^2} \text{ m}$ Since $\pi^2 \approx 9.87$ , $L \approx \frac{9.8}{9.87} \approx 0.993 \text{ m}$

Final Answer The length of a seconds pendulum is approximately 1 m.

OscillationsClass 11 Physics Notes

Introduction

Periodic and Oscillatory Motions

Period and frequency

Given

To Find

Formula

Solution

Displacement

Solution

Simple Harmonic Motion

Solution

Simple Harmonic Motion and Uniform Circular Motion

Given

To Find

Formula

Solution

Velocity and Acceleration in Simple Harmonic Motion

Velocity

Acceleration

Given

To Find

Formula

Solution

Force Law for Simple Harmonic Motion

Given

To Find

Solution

Energy in Simple Harmonic Motion

Kinetic Energy (K)

Potential Energy (U)

Total Energy (E)

Given

To Find

Formula

Solution

The Simple Pendulum

Given

To Find

Formula

Solution