NCERT Solutions Class 11 Physics Chapter 13 – Oscillations

NCERT Solutions for Class 11 Physics Chapter 13 Oscillations explain periodic motion and Simple Harmonic Motion SHM, where motion repeats at regular intervals. The chapter covers important concepts such as amplitude, time period, frequency, phase, energy in SHM, spring oscillations, and simple pendulum.

NCERT Solutions for Class 11 Physics Chapter 13 by ALLEN provide clear, step-by-step explanations of SHM using trigonometric functions and phasors. The solutions cover time period derivations, energy changes, and numerical problems for CBSE, JEE, and NEET preparation.

1.0Class 11 Physics Chapter 13 Oscillations: Key Concepts

This chapter examines the dynamics of oscillating systems and the energy conversions that sustain them. Key lessons include:

Periodic and Oscillatory Motion: Differentiating between motion that repeats (Periodic) and motion that moves to-and-fro about a mean position (Oscillatory).
Simple Harmonic Motion (SHM): Defining the motion where F = -kx or $a = - ω^{2} x$
Displacement: $x (t) = A cos (ω t + ϕ)$
Velocity: $v (t) = - ω A sin (ω t + ϕ)$
Acceleration: $a (t) = - ω^{2} A cos (ω t + ϕ)$
Energy in SHM: Analyzing the exchange between Potential Energy ( $U = \frac{1}{2} k x^{2}$ ) and Kinetic Energy ( $K = \frac{1}{2} m v^{2}$ ).
Total Energy (E = K + U) remains constant in an ideal system.
The Simple Pendulum: Deriving the time period for a point mass suspended by a string:
$T = 2 π \frac{L}{g}$
Oscillations due to a Spring: Mastering the time period for a block-spring system: $T = 2 π \frac{m}{k}$ .
Damped Simple Harmonic Motion: Understanding how real-world friction and air resistance cause the amplitude to decrease over time.
Forced Oscillations and Resonance: Studying systems driven by an external periodic force and the phenomenon of Resonance, where the driving frequency matches the natural frequency.

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 13

SOLVED EXAMPLES

On an average, a human heart is found to beat 75 times in a minute. Calculate its frequency and period. Sol. The beat frequency of heart, $f = \frac{75}{( 1 min )}$

$= 75/ (60 s) = 1.25 s^{- 1} = 1.25 Hz$

The time period, $T = \frac{1}{f} = \frac{1}{( 1.25 s ^{- 1} )} = 0.8 s$

Which of the following functions of time represent (a) periodic and (b) non-periodic motion? Give the period for each case of periodic motion [ $ω$ is any positive constant]. (1) $sin ω t + cos ω t$ (ii) $sin ω t + cos 2 ω t + sin 4 ω t$ (iii) $e^{- ω t}$ (iv) $lo g (ω t)$ Sol. (i) $sin ω t + cos ω t$ is a periodic function, it can also be written as

$sin ω t + cos ω t = 2 sin (ω t + \frac{π}{4}) = 2 sin (ω t + \frac{π}{4} + 2 π) = 2 sin [ω (t + \frac{2 π}{ω}) + \frac{π}{4}]$

The periodic time of the function is $\frac{2 π}{ω}$ . (ii) This is an example of a periodic motion. It can be noted that each term represents a periodic function with a different angular frequency. Since period is the least interval of time after which a function repeats its value, sinct has a period $T_{0} = \frac{2 π}{ω}; cos 2 ω t$ has a period $\frac{π}{ω} = \frac{T _{0}}{2}$ ; and $sin 4 ω t$ has a period

$2 π /4 ω = \frac{T _{0}}{4} .$

The period of the first term is a multiple of the periods of the last two terms. Therefore, the smallest interval of time after which the sum of the three terms repeats is $T_{0}$ , and thus, the sum is a periodic function with a period $\frac{2 π}{ω}$ . (iii) The function $e^{- ω t}$ is not periodic, it decreases monotonically with increasing time and tends to zero as $t \to \infty$ and thus, never repeats its value. (iv) The function $lo g (ω t)$ increases monotonically with time t. It, therefore, never repeats its value and is a non periodic function. It may be noted that as $t \to \infty, lo g (ω t)$ diverges to $\infty$ . It, therefore, cannot represent any kind of physical displacement.

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. (a) $sin ω t - cos ω t$ (b) $sin^{2} ω t$ Sol. (a) $sin ω t - cos ω t = sin ω t - sin (\frac{π}{2} - ω t)$

$= 2 cos (\frac{π}{4}) sin (ω t - \frac{π}{4}) = 2 sin (ω t - \frac{π}{4})$

This function represents a simple harmonic motion having a period

$T = \frac{2 π}{ω} and a phase angle (\frac{- π}{4}) or (\frac{7 π}{4})$

(b) $sin^{2} ω t = \frac{1}{2} - \frac{1}{2} cos 2 ω t$ The function is periodic having a period $T = \frac{π}{ω}$ . It also represents a harmonic motion with the point of equilibrium occurring at $\frac{1}{2}$ instead of zero.

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.

Sol. (a) At $t = 0, OP$ makes an angle of $4 5^{\circ} = \frac{π}{4}$ rad with the (positive direction of) x-axis. After time t, it covers an angle $\frac{2 π}{T} t$ in the anticlockwise sense, and makes an angle of $\frac{2 π}{T} t + \frac{π}{4}$ with the x-axis. The projection of OP on the x-axis at time t is given by,

$x (t) = A cos (\frac{2 π}{T} t + \frac{π}{4})$

For $T = 4 s$ ,

$x (t) = A cos (\frac{2 π}{4} t + \frac{π}{4})$

which is SHM of amplitude A, period 4s, and an initial phase $= \frac{π}{4}$ (b) In this case at $t = 0, OP$ makes an angle of $9 0^{\circ} = \frac{π}{2}$ with the x -axis. After a time t , it covers an angle of $\frac{2 π}{T} t$ in the clockwise sense and makes an angle of $(\frac{π}{2} - \frac{2 π}{T} t)$ with the x -axis. The projection of OP on the x-axis at time t is given by

$x (t) = B cos (\frac{π}{2} - \frac{2 π}{T} t) = B sin (\frac{2 π}{T} t)$

For $T = 30 s$ ,

$x (t) = B sin (\frac{π}{15} t)$

Writing this as $x (t) = B cos (\frac{π}{15} t - \frac{π}{2})$ , and comparing with Eq. $x (t) = A cos (ω t$ $+ Φ)$ . We find that this represents SHM of amplitude B, period 30s, and an initial phase of $- \frac{π}{2}$ .

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

$x = 5 cos [2 π t + \frac{π}{4}] .$

At $t = 1.5 s$ , calculate the (a) displacement, (b) speed and (c) acceleration of the body. Sol. The angular frequency of the body, $ω = 2 π rads^{- 1}$ and its time period $T = 1 s$ . At $t = 1.5 s$ (a) displacement

$= (5.0 m) cos [(2 π \times 1.5) + \frac{π}{4}] = (5.0 m) cos [(3 π + \frac{π}{4})] = - 5.0 \times 0.707 m = - 3.535 m$

(b) The speed of the body is given by,

$v = \frac{dx}{dt} \Rightarrow v = \frac{d}{dt} [5 cos (2 π t + \frac{π}{4})] v = - 5 \times 2 π sin (2 π t + \frac{π}{4}) = - 10 π sin (2 π \times 1.5 + \frac{π}{4}) = - 10 π \times 0.707 ms^{- 1} v = - 22 ms^{- 1}$

$a = - ω^{2} x = - (2 π)^{2} \times displacement = - (2 π)^{2} \times (- 3.535) = 140 ms^{- 2}$

Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shown in 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.

Sol. . Let the mass be displaced by a small distance x to the right side of the equilibrium position, as shown in Figure. Under this situation the spring on the left side gets elongated by a length equal to x and that on the right side gets compressed by the same length. The forces acting on the mass are then,

F1 = –kx (force exerted by the spring on the left side, trying to pull the mass towards the mean position) F2 = –kx (force exerted by the spring on the right side, trying to push the mass towards the mean position)

The net force, F, acting on the mass is then given by,

F = –2kx

Hence the force acting on the mass is proportional to the displacement and is directed towards the mean position; therefore, the motion executed by the mass is simple harmonic. The time period of oscillations is,

$T = 2 π \frac{m}{2 k}$

A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of $50 N m^{- 1}$ . 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. Sol. The block executes SHM, its angular frequency, as given by,

$ω = \frac{k}{m} = \frac{50 Nm ^{- 1}}{1 kg} = 7.07 rad s^{- 1}$

Its displacement at any time t is then given by,

$x (t) = 0.1 cos (7.07 t)$

Therefore, when the particle is 5cm away from the mean position, we have

$0.05 = 0.1 cos (7.07 t)$

or $cos (7.07 t) = 0.5$ and hence

$sin (7.07 t) = \frac{3}{2} = 0.866$

Then, the velocity of the block at $x = 5 cm$ is

$v = \frac{d}{dt} [x (t)] \Rightarrow v = 0.1 \times 7.07 [- sin (7.07 t)] v = - 0.1 \times 7.07 \times 0.866 v = - 0.61 m / s$

Hence the K.E. of the block,

$K.E. = \frac{1}{2} m v^{2} = \frac{1}{2} \times 1 \times (- 0.61)^{2} = 0.19 J$

The P.E. of the block,

$P.E. = \frac{1}{2} kx^{2} = \frac{1}{2} (50 N m^{- 1} \times 0.05 m \times 0.05 m) = 0.0625 J$

The total energy of the block at $x = 5 cm$ ,

$= K.E. + P.E. = 0.25 J$

we also know that at maximum displacement, K.E. is zero and hence the total energy of the system is equal to the P.E. Therefore, the total energy of the system,

$= \frac{1}{2} (50 N m^{- 1} \times 0.1 m \times 0.1 m) = 0.25 J$

which is same as the sum of the two energies at a displacement of 5 cm . This is in conformity with the principle of conservation of energy.

What is the length of a simple pendulum, which ticks seconds ? Sol. The time period of a simple pendulum is given by,

$T = 2 π \frac{L}{g}$

From this relation one gets,

$L = \frac{gT ^{2}}{4 π ^{2}}$

The time period of a simple pendulum, which ticks seconds, is 2 s. Therefore, for $g = 9.8 ms^{- 2}$ and $T = 2 s, L$ is

$= \frac{9.8 ( ms ^{- 2} ) \times 4 ( s ^{2} )}{4 π ^{2}} = 1 m$

EXERCISE QUESTION WITH SOLUTIONS

Which of the following examples represent periodic motion? (a) A swimmer completing one (return) trip from one bank of a river to the other and back. (b) A freely suspended bar magnet displaced from its N-S direction and released. (c) A hydrogen molecule rotating about its centre of mass. (d) An arrow released from a bow. Sol. (a) The swimmer's motion is not periodic. Though the motion of a swimmer is to and fro but will not have a definite period. (b) The motion of a freely-suspended magnet, if displaced from its N-S direction and released, is periodic because the magnet oscillates about its position with a definite period of time. (c) When a hydrogen molecule rotates about its centre of mass, it comes to the same position again and again after an equal interval of time. Such a motion is periodic. (d) An arrow released from a bow moves only in the forward direction. It does not come backward. Hence, this motion is not a periodic.
Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion? (a) the rotation of earth about its axis. (b) motion of an oscillating mercury column in a U-tube. (c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point. (d) general vibrations of a polyatomic molecule about its equilibrium position.

Sol. (a) It is periodic but not simple harmonic motion because it is not to and fro about a fixed point. (b) It is a simple harmonic motion because the mercury moves to and fro on the same path, about the fixed position, with a certain period of time. (c) It is simple harmonic motion because the ball moves to and fro about the lowermost point of the bowl when released. Also, the ball comes back to its initial position in the same period of time, again and again. (d) A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the superposition of individual simple harmonic motions of a number of different molecules. Hence, it is not simple harmonic, but periodic.

Figure depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion) ?

exercise-ques-3-(a)-class-11-physics-chap-13

exercise-ques-3-(b)-class-11-chap-13-physics

physics-class-11-chap-13-exercise-ques-3-(c)

physics-exercise-ques-3-(d)-class-11-chap-13

Sol. (a) It is not a periodic motion. This represents a unidirectional, linear uniform motion. There is no repetition of motion in this case. (b) In this case, the motion of the particle repeats itself after 2s. Hence, it is a periodic motion, having a period of 2s. (c) It is not a periodic motion. This is because the particle repeats the motion in one position only. For a periodic motion, the entire motion of the particle must be repeated in equal intervals of time. (d) In this case, the motion of the particle repeats itself after 2s. Hence, it is a periodic motion, having a period of 2s.

Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion ( $ω$ is any positive constant): (a) $sin ω t - cos ω t$ (b) $sin^{3} ω t$ (c) $3 cos (\frac{π}{4} - 2 ω t)$ (d) $cos ω t + cos 3 ω t + cos 5 ω t$ (e) $exp (- ω^{2} t^{2})$ Sol. (a) SHM The given function is

$sin ω t - cos ω t = 2 [\frac{1}{2} sin ω t - \frac{1}{2} cos ω t] = 2 [sin ω t \times cos \frac{π}{4} - cos ω t \times sin \frac{π}{4}] = 2 sin (ω t - \frac{π}{4})$

This function represent SHM as it can be written in the form: $a sin (ω t + Φ)$ Its period is : $\frac{2 π}{ω}$

(b) Periodic but not SHM The given function is

$sin^{3} ω t = \frac{1}{4} [3 sin ω t - sin 3 ω t]$

The terms $sin ω t$ and $sin 3 ω t$ individually represent simple harmonic motion (SHM). However, the superposition of these two SHM is periodic and not simple harmonic. Its period is : $\frac{2 π}{ω}$ (c) SHM The given function is : $3 cos [\frac{π}{4} - 2 ω t]$

$= 3 cos [2 ω t - \frac{π}{4}]$

This function represents simple harmonic motion because it can be written in the form: $a cos (ω t + Φ)$ Its period is : $\frac{2 π}{2 ω} = \frac{π}{ω}$ (d) Periodic, but not SHM The given function is,

$cos ω t + cos 3 ω t + cos 5 ω t .$

Each individual cosine function represents SHM. However, the superposition of these three simple harmonic motions is periodic, but not simple harmonic. (e) Non-periodic motion The given function $exp (- ω^{2} t^{2})$ is an exponential function. Exponential functions do not repeat themselves. Therefore, it is a non-periodic motion. (f) The given function $1 + ω t + ω^{2} t^{2}$ is non-periodic.

A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is (a) at the end A. (b) at the end B. (c) at the mid-point of AB going towards A, (d) at 2 cm away from B going towards A, (e) at 3 cm away from A going towards B, and (f) at 4 cm away from B going towards A. Sol. From figure, where A and B represent the two extreme positions of a SHM.

exercise-ques-5-physics-chap-13-class-11

For velocity, the direction from A to B is taken positive. The acceleration and the force, along AP are taken as positive and along BP are taken as negative. (a) At the end A, the particle executing SHM is momentarily at rest being its extreme position of motion. Therefore, its velocity is zero. Acceleration is positive because it is directed along AP, Force is also Positive since the force is directed along AP. (b) At the end B, velocity is zero. Here, acceleration and force are negative as they are directed along BP. (c) At the mid point of AB going towards A, the particle is at its mean position P , with a tendency to move along PA. Hence, velocity is negative. Both acceleration and force are zero.

(d) At 2 cm away from B going towards A, the particle is at Q, with a tendency to move along QP, which is negative direction. Therefore, velocity, acceleration and force all are negative. (e) At 3 cm away from A going towards B, the particle is at R, with a tendency to move along RP, which is positive direction. Here, velocity, acceleration and force all are positive. (f) At 4 cm away from A going towards A, the particle is at S, with a tendency to move along SA, which is negative direction. Therefore, velocity is negative but acceleration is directed towards mean position, along SP. Hence it is positive and also force is positive similarly.

Which of the following relationships between the acceleration 'a' and the displacement 'x' of a particle involve simple harmonic motion? (a) $a = 0.7 x$ (b) $a = - 200 x^{2}$ (c) $a = - 10 x$ (d) $a = 100 x$ Sol. (c) In SHM, acceleration a is related to displacement by the relation of the form $a = - kx$ , for relation (c).7
The motion of a particle executing simple harmonic motion is described by the displacement function.

$x (t) = A cos (ω t + Φ) .$

If the initial $(t = 0)$ position of the particle is 1 cm and its initial velocity is $ω cm / s$ , what are its amplitude and initial phase angle? The angular frequency of the particle is $π rad s^{- 1}$ .If instead of the cosine function, we choose the sine function to describe the SHM : $x = B sin (ω t + α)$ , what are the amplitude and initial phase of the particle with the above initial conditions. Sol. Initially, at $t = 0$ : Displacement, $x = 1 cm$ Initial velocity, $v = ω cm / sec$ . Angular frequency, $ω = π rads^{- 1}$ . It is given that,

$x (t) = A cos (ω t + Φ) 1 = A cos (ω \times 0 + Φ) = A cos Φ A cos Φ = 1 \dots .. (i)$

Velocity, $v = \frac{dx}{dt}$

$ω = - A ω sin (ω t + Φ) 1 = - A sin (ω \times 0 + Φ) = - A sin Φ A sin Φ = - 1 \dots .. (ii)$

Squaring and adding equations (i) and (ii), we get:

$A^{2} (sin^{2} Φ + cos^{2} Φ) = 1 + 1 A^{2} = 2 ∴ A = 2 cm$

Dividing equation (ii) by equation (i), we get: $tan Φ = - 1$

$∴ Φ = \frac{3 π}{4}, \frac{7 π}{4}, \dots\dots$

SHM is given as:

$x = B sin (ω t + α)$

Putting the given values in this equation, we get:

$1 = B sin [ω \times 0 + α]$

B $sin α = 1$ Velocity, $v = ω B cos (ω t + α)$ Substituting the given values, we get:

$π = π B sin α B sin α = 1$

Squaring and adding equations (iii) and (iv), we get:

$B^{2} [sin^{2} α + cos^{2} α] = 1 + 1 B^{2} = 2 ∴ B = 2 cm$

Dividing equation (iii) by equation (iv), we get:

$\frac{B s i n α}{B c o s α} = \frac{1}{1}$

$tan α = 1 = tan \frac{π}{4} ∴ α = \frac{π}{4}, \frac{5 π}{4} \dots ..$

A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s . What is the weight of the body?

Sol. Maximum mass that the scale can read,

$M = 50 kg$

Maximum displacement of the spring = Length of the scale, $l = 20 cm = 0.2 m$ Time period, $T = 0.6 s$ Maximum force exerted on the spring,

$F = Mg$

where,

$g = acceleration due to gravity = 9.8 m / s^{2} F = 50 \times 9.8 = 490$

∴ Spring constant,

$k = \frac{F}{l} = \frac{490}{0.2} = 2450 N m^{- 1} .$

Mass m, is suspended from the balance. Time period, $T = 2 π \frac{m}{k}$

$∴ m = (\frac{T}{2 π})^{2} \times k = (\frac{0.6}{2 \times 3.14})^{2} \times 2450 = 22.36 kg$

∴ Weight of the body $= mg = 22.36 \times 9.8$

$= 219.167 N$

Hence, the weight of the body is about 219N.

A spring with a spring constant $1200 N m^{- 1}$ is mounted on a horizontal table as shown in Figure. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.

class-11-exercise-ques-9-physics-chap-13

Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass. Sol. Spring constant, $k = 1200 N m^{- 1}$ Mass, $m = 3 kg$ Displacement, $A = 2.0 cm = 0.02 m$ (i) Frequency of oscillation $v$ , is given by the relation

$v = \frac{1}{T} = \frac{1}{2 π} \frac{k}{m}$

Where, T is time period

$∴ v = \frac{1}{2 \times 3.14} \frac{1200}{3} = 3.18 m / s$

Hence, the frequency of oscillation is 3.18 cycles per second (ii) Maximum acceleration (a) is given by the relation:

$a = ω^{2} A$

where,

$ω = Angular frequency = \frac{k}{m}$

A = maximum displacement

$∴ a = \frac{k}{m} A = \frac{1200 \times 0.02}{3} = 8 ms^{- 2}$

Hence, the maximum acceleration of the mass is $8.0 m / s^{2}$ .

(iii) Maximum velocity, $v_{max} = A ω$ .

$= A \frac{k}{m} = 0.02 \times \frac{1200}{3} = 0.4 m / s .$

Hence, the maximum velocity of the mass is 0.4 m/s.

In Exercise 9, let us take the position of mass when the spring is unstreched as $x = 0$ , and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch $(t = 0)$ , the mass is (a) at the mean position, (b) at the maximum streched position, and (c) at the maximum compressed position.

In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase? Sol. Distance travelled by the mass sideways, $a = 2.0 cm$ Angular frequency of oscillation:

$ω = \frac{k}{m} = \frac{1200}{3} = 400 = 20 rad s^{- 1}$

(a) As time is noted from the mean position, hence using $x =$ asin $ω t$ , we have $x = 2 sin 20 t$ (b) At maximum stretched position, the body is at the extreme right position, with an initial phase of $\frac{π}{2} rad$ . Then, $x = 2 sin (ω t + \frac{π}{2}) = 2 cos ω t = 2 cos 20 t$ (c) At maximum compressed position, the body is at left position, with an initial phase of $3 π /2 rad$ . Then,

$x = 2 sin (ω t + \frac{3 π}{2}) = - 2 cos ω t = - 2 cos 20 t$

The functions neither differ in amplitude nor in frequency. They differ in initial phase.

Figures correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure.

physics-class-11-chap-13-exercise-ques-11

Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case. Sol. (a) Time period, $T = 2 s$ Amplitude, $A = 3 cm$ At time, $t = 0$ , the radius vector OP makes an angle $π /2$ with the positive x-axis, i.e., phase angle $Φ = + π /2$ Therefore, the equation of simple harmonic motion for the x-projection of OP, at the time t, is given by the displacement equation:

$x = A cos [\frac{2 π t}{T} + ϕ] = 3 cos (\frac{2 π t}{2} + \frac{π}{2}) = - 3 sin (\frac{2 π t}{2}) ∴ x = - 3 sin π t cm$

(b) Time Period, $t = 4 s$ Amplitude, $A = 2 m$ At time $t = 0, OP$ makes an angle $π$ with the x-axis, in the anticlockwise direction, Hence, phase angle $Φ = + π$ Therefore, the equation of simple harmonic motion for the x-projection of OP, at the time t, is given as:

$x = 2 cos [\frac{2 π t}{T} + ϕ] = 2 cos (\frac{2 π t}{4} + π) ∴ x = - 2 cos (\frac{π}{2} t) m$

Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial ( $t = 0$ ) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: ( x is in cm and t is in s). (a) $x = - 2 sin (3 t + \frac{π}{3})$ (b) $x = cos (\frac{π}{6} - t)$ (c) $x = 3 sin (2 π t + \frac{π}{4})$ (d) $x = 2 cos π t$ Sol. (a)

$x = - 2 sin (3 t + \frac{π}{3}) = 2 cos (3 t + \frac{π}{3} + \frac{π}{2}) = 2 cos (3 t + \frac{5 π}{6})$

If this equation is compared with the standard SHM equation

$x = A cos (\frac{2 π}{T} t + ϕ), then we get:$

Amplitude, $A = 2 cm$

$Phase angle, Φ = \frac{5 π}{6} = 15 0^{\circ}$

Angular velocity $= ω = \frac{2 π}{T} = 3 rad / sec$ .

exercise-ques-12-(a)-class-11-chap-13-physics

The motion of the particle can be plotted as shown in figure.

(b) $x = cos (\frac{π}{6} - t) = cos (t - \frac{π}{6})$ If this equation is compared with the standard SHM equation $x = A cos (\frac{2 π}{T} t + ϕ)$ , then we get: Amplitude, $A = 1$ Phase angle, $Φ = - \frac{π}{6} = - 3 0^{\circ}$ . Angular velocity, $ω = \frac{2 π}{T} = 1 rad / s$ .

physics-class-11-chap-13-exercise-ques-3-(b)

The motion of the particle can be plotted as shown in figure.

(c)

$x = 3 sin (2 π t + \frac{π}{4}) = - 3 cos [(2 π t + \frac{π}{4}) + \frac{π}{2}] = - 3 cos (2 π t + \frac{3 π}{4})$

If this equation is compared with the standard SHM equation $x = A cos (\frac{2 π}{T} t + ϕ)$ , then we get: Amplitude, $A = 3 cm$ Phase angle, $Φ = \frac{3 π}{4} = 13 5^{\circ}$ Angular velocity, $ω = \frac{2 π}{T} = 2 rad / s$

chap-13-physics-exercise-ques-12-(c)-class-11

The motion of the particle can be plotted as shown in figure.

(d) $x = 2 cos π t$ If this equation is compared with the standard SHM equation $x = A cos (\frac{2 π}{T} t + ϕ)$ , then we get: Amplitude, $A = 2 cm$ Phase angle, $Φ = 0$ Angular velocity, $ω = π rad / s$ .

exercise-ques-12-(d)-class-11-physics-chap-13

The motion of the particle can be plotted as shown in figure.

Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. (b) is stretched by the same force F.

physics-exercise-ques-13-chap-13-class-11

(a) What is the maximum extension of the spring in the two cases? (b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case? Sol. (a) The maximum extension of the spring in both cases will = Flk, where k is the spring constant of the springs used. (b) In Fig.(a) if x is the extension in the spring, when mass m is returning to its mean position after being released free, then restoring force on the mass is $F = - kx$ , i.e., $F \propto - x$ As, this F is directed towards mean position of the mass, hence the mass attached to the spring will execute SHM. Spring factor = spring constant = k inertia factor = mass of the given block mass of block $= m$ As time period, $T = 2 π \frac{inertia factor}{spring factor}$ $∴ T = 2 π \frac{m}{k}$ In figure (b), we have a two body system of spring constant k and reduced mass $μ$ ,

$μ = \frac{m \times m}{m + m} = \frac{m}{2}$

Inertia factor $= \frac{m}{2}$ Spring factor $= k$

$∴ time period, T = 2 π \frac{m /2}{k} = 2 π \frac{m}{2 k}$

The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m . If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?

Sol. Angular frequency of the piston, $ω = 200 rad / min$ . Stroke $= 1.0 m$ Amplitude, $A = \frac{1.0}{2} = 0.5 m$ The maximum speed $(V_{max})$ of piston is given by the relation: $v_{max} = A ω = 200 \times 0.5 = 100 m / min$ .

The acceleration due to gravity on the surface of moon is $1.7 ms^{- 2}$ . What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is $9.8 ms^{- 2}$ ) Sol. Acceleration due to gravity on the surface of moon, $g^{'} = 1.7 m s^{2}$ Acceleration due to gravity on the surface of earth, $g = 9.8 ms^{- 2}$ Time period of a simple pendulum on earth, $T = 3.5 s$ $T = 2 π \frac{l}{g}$ Where, $l$ is the length of the pendulum

$∴ l = \frac{T ^{2}}{( 2 π ) ^{2}} \times g = \frac{( 3.5 ) ^{2}}{4 \times ( 3.14 ) ^{2}} \times 9.8 m = 3.04 m$

The length of pendulum remains constant On moon's surface, time period,

$T^{'} = 2 π \frac{l}{g ^{'}} = 2 π \frac{3.04}{1.7} = 8.4 s$

Hence, the time period of the simple pendulum on the surface of moon is 8.4s.

A simple pendulum of length $l$ and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period? Sol. The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car. Acceleration due to gravity $= g$ Centripetal acceleration $= \frac{v ^{2}}{R}$ where, v is the uniform speed of the car and R is the radius of the track Effective acceleration (g'), is given as:

$g^{'} = g^{2} + \frac{v ^{4}}{R ^{2}}$

∴ Time period, $T = 2 π \frac{l}{g ^{'}}$ ,

$= 2 π \frac{l}{g ^{2} + \frac{v ^{4}}{R ^{2}}}$

Cylindrical piece of cork of density of base area A and height h floats in a liquid of density $ρ_{1}$ . The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period

$T = \frac{h ρ}{ρ _{1} g}$

where $ρ$ is the density of cork. (Ignore damping due to viscosity of the liquid). Sol. Base area of the cork $= A$ Height of the cork = h Density of the liquid $= ρ_{1}$ Density of the cork $= ρ$ In equilibrium: Weight of the cork = Weight of the liquid displaced by the floating cork Let the cork be depressed slightly by x. As a result, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork.

Up-thrust = Restoring force Restoring force, $F =$ Weight of the extra water displaced

$F = - (Volume \times Density \times g)$

Volume = Area × Distance through which the cork is depressed

$Volume = Ax ∴ F = - Ax ρ_{1} g$

According to the force law:

$F = - kx \Rightarrow k = - \frac{F}{x} k = - \frac{( - Ax ρ _{1} g )}{x} where, k is constant k = A ρ_{1} g$

The time period of the oscillations of the cork:

$T = 2 π \frac{m}{k}$

where, $m =$ Mass of the cork

$m = Volume of the cork \times Density = Base area of the cork \times height of the cork \times Density of the cork = Ah ρ$

Hence, the expression for the time period becomes:

$T = 2 π \frac{Ah ρ}{A ρ _{1} g} = 2 π \frac{h ρ}{ρ _{1} g}$

One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.

Sol. Area of cross-section of the U-tube = A Density of the mercury column $= ρ$ Acceleration due to gravity $= g$ Restoring force, $F =$ Weight of the mercury column of a certain height

$F = - (Volume \times Density \times g) F = - (A \times 2 h \times ρ \times g) = - 2 A ρ g h$

Where, 2h is the height of the mercury column in the two arms. k is a constant, given by

$k = - \frac{F}{h} = 2 A ρ g$

Time period, $T = 2 π \frac{m}{k} = 2 π \frac{m}{2 A ρ g}$ where, m is the mass of the mercury column

Let $l$ be the length of the total mercury in the U-tube

Mass of mercury, $m =$ Volume of mercury × Density of mercury $= A lρ$

$∴ T = 2 π \frac{A lρ}{2 A ρ g} = 2 π \frac{l}{2 g}$

Hence, the mercury column executes simple harmonic motion with time period $2 π \frac{l}{2 g}$ .

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

Prepare with NCERT Solutions for Class 11 Physics, chapter-wise, with detailed explanations for each question in the textbook. Learn important concepts, understand good steps to solve the problems and enhance your understanding of Physics basics.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Units and Measurements
Chapter 2	Motion in a Straight Line
Chapter 3	Motion in a Plane
Chapter 4	Laws of Motion
Chapter 5	Work, Energy and Power
Chapter 6	System of Particles and Rotational Motion
Chapter 7	Gravitation
Chapter 8	Mechanical Properties of Solids
Chapter 9	Mechanical Properties of Fluids
Chapter 10	Thermal Properties of Matter
Chapter 11	Thermodynamics
Chapter 12	Kinetic Theory
Chapter 14	Waves

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 13

Graphical Analysis: Understand x-t, v-t, and a-t graphs and phase differences, including why velocity leads displacement by $π /2$
Phase and Phasors: Learn about initial phase $(ϕ)$ and the use of rotating vectors or phasors to solve SHM problems.
Energy Conservation: Understand why the total energy in SHM is proportional to the square of amplitude, $E \propto A^{2}$
Numerical Solutions: Solve problems involving springs in series and parallel and pendulums in accelerating frames with step-by-step calculations.
Prepared by ALLEN Subject Experts: Get solutions prepared by ALLEN subject experts with conceptual clarity and alignment with the latest NCERT syllabus for board and competitive exams.

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NCERT Solutions Class 11 Physics Chapter 13 – Oscillations

1.0Class 11 Physics Chapter 13 Oscillations: Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 13

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3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 13

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