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NCERT Solutions for Class 11 Physics Chapter 14 Oscillations

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations

Edited By Vishal kumar | Updated on Sep 07, 2023 12:02 PM IST

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations hold great significance in the Class 11 curriculum. On this Careers360 page of NCERT solutions , you'll find comprehensive ch 14 physics class 11 ncert solutions crafted by experts, presented in a clear and straightforward manner. This resource covers a total of 25 questions, encompassing exercises from 14.1 to 14.19, and 14.20 to 14.25 in the additional exercise section. The oscillation exercise class 12 solutions are conveniently available in PDF format, allowing students to access and download them for free whenever they require assistance.

This Story also Contains
  1. NCERT Solutions for Class 11 Physics Chapter 14: Oscillations
  2. NCERT Solutions For Class 11 Physics Chapter 14 Oscillations: Exercise Solution
  3. NCERT solutions for class 11 physics chapter 14 oscillations: Additional Exercise Solution
  4. oscillation class 11 physics
  5. NCERT solutions for class 11 physics chapter-wise
  6. Oscillations Class 11 NCERT Topics
  7. Importance of NCERT Solutions for Class 11 Physics Chapter 14 Oscillations:
  8. Key features of NCERT Solutions for Class 11 Physics Chapter 14
  9. Subject wise NCERT Exemplar solutions

Problems with motion along a straight line, motion in a plane, projectile motion, etc. were discussed in the previous chapters. NCERT Physics chapter 14 Oscillations Class 11 solutions explains problems on periodic and oscillatory motions. The motion which repeats after a certain interval of time is called periodic motion. For example, the motion of a planet around the sun, the motion of pendulum of a wall clock etc are periodic. To and fro periodic motion about a mean position is known as oscillatory motion.

Background wave

CBSE NCERT solutions for Class 11 Physics chapter 14 Oscillations have questions on simple harmonic motion (SHM). SHM is the simplest form of oscillatory motion. In SHM the force on the oscillating body is directly proportional to the displacement about the mean position and is directed towards the mean position. Oscillation chapter class 11 solution are an important tool to score well in the exams and also they are useful if you want to study other subjects of other classes as well. The concept studied in NCERT becomes easy to understand with the help of NCERT solutions for class 11.

This chapter has been renumbered as Chapter 13 in accordance with the CBSE Syllabus 2023–24.

Free download ch 14 physics class 11 ncert solutions PDF for CBSE exam.

NCERT Solutions for Class 11 Physics Chapter 14: Oscillations

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NCERT Solutions For Class 11 Physics Chapter 14 Oscillations: Exercise Solution

Q. 14.1 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 bank.

(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

Answer:

(a) The motion is not periodic though it is to and fro.

(b) The motion is periodic.

(c) The motion is periodic.

(d) The motion is not periodic.

Q. 14.2 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.

Answer:

(a) Periodic but not S.H.M.

(b) S.H.M.

(c) S.H.M.

(d) Periodic but not S.H.M.M [A polyatomic molecule has a number of natural frequencies, so its vibration is a superposition of SHM’s of a number of different frequencies. This is periodic but not SHM]

Q .14.3 Fig. 14.23 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)?

1650449930218

Answer:

The x-t plots for linear motion of a particle in Fig. 14.23 (b) and (d) represent periodic motion with both having a period of motion of two seconds.

Q. 14.4 (a) 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) sinωtcosωt

Answer:

sinωtcosωt=2(12sinωt12cosωt)=2(cosπ4sinωtsinπ4cosωt)=2sin(ωtπ4)

Since the above function is of form Asin(ωt+ϕ) it represents SHM with a time period of 2πω

Q. 14.4 (b) 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):

(b) sin3ωt

Answer:

sin3ωt=3sinωt4sin3ωtsin3ωt=14(3sinωtsin3ωt)

The two functions individually represent SHM but their superposition does not give rise to SHM but the motion will definitely be periodic with a period of 2πω

Q. 14.4 (c) 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):

(c) 3cos(π/42ωt)

Answer:

The function represents SHM with a period of πω

Q. 14.4 (c)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 ( is any positive constant): (c)

Q.14.4 (d) 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):

(d) cosωt+cos3ωt+cos5ωt

Answer:

Here each individual functions are SHM. But superposition is not SHM. The function represents periodic motion but not SHM.

period=LCM(2πω,2π3ω,2π5ω)=2πω

Q. 14.5 (c) A particle is in linear simple harmonic motion between two points, A and B , 10cm 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

(c) at the mid-point of AB going towards A ,

Answer:

Velocity is negative that is towards A and its magnitude is maximum. Acceleration and force are zero.

Q. 14.6 Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?

(a) a=0.7x

(b) a=200x2

(c) a=10x

(d) a=100x3

Answer:

Only the relation given in (c) represents simple harmonic motion as the acceleration is proportional in magnitude to the displacement from the midpoint and its direction is opposite to that of the displacement from the mean position.

Q. 14.7 The motion of a particle executing simple harmonic motion is described by the displacement function, x(t)=Acos(ωt+ϕ).

If the initial (t=0) position of the particle is 1cm and its initial velocity is ωcm/s, what are its amplitude and initial phase angle ? The angular frequency of the particle is πs1. If instead of the cosine function, we choose the sine function to describe the SHM : x=Bsin(ωt+α), what are the amplitude and initial phase of the particle with the above initial conditions.

Answer:

ω=π rad s1

x(t)=Acos(πt+ϕ)

at t = 0

x(0)=Acos(π×0+ϕ)1=Acosϕ               (i)

v=dx(t)dtv(t)=Aπsin(πt+ϕ)

at t = 0

v(0)=Aπsin(π×0+ϕ)ω=Aπsinϕ 1=Asinϕ                      (ii)

Squaring and adding equation (i) and (ii) we get

12+12=(Acosϕ)2+(Asinϕ)22=A2cos2ϕ+A2sin2ϕ2=A2A=2

Dividing equation (ii) by (i) we get

tanϕ=1ϕ=3π4,7π4,11π4......

x(t)=Bsin(πt+α)

at t = 0

x(0)=Bsin(π×0+α)1=Bsinα               (iii)

v=dx(t)dtv(t)=Bπcos(πt+α)

at t = 0

v(0)=Bπcos(π×0+α)ω=Bπcosα 1=Bcosα                      (iv)

Squaring and adding equation (iii) and (iv) we get

12+12=(Bsinα)2+(Bcosα)22=B2sin2α+B2cos2α2=B2B=2

Dividing equation (iii) by (iv) we get

tanα=1α=π4,5π4,9π4......

Q. 14.8 A spring balance has a scale that reads from 0 to 50kg. The length of the scale is 20cm, A body suspended from this balance, when displaced and released, oscillates with a period of 0.6s. What is the weight of the body?

Answer:

Spring constant of the spring is given by

k=Weight of Maximum mass the scale can readMaximum displacement of the scalek=50×9.820×102k=2450 Nm1

The time period of a spring attached to a body of mass m is given by

T=2πmkm=T2k4π2m=(0.6)2×24504π2m=22.34 kgw=mgw=22.34×9.8w=218.95 N

Q. 14.9 (i) A spring having with a spring constant 1200Nm1 is mounted on a horizontal table as shown in Fig. 14.24. A mass of 3kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0cm and released.

1424

Determine

(i) the frequency of oscillations,

Answer:

The frequency of oscillation of an object of mass m attached to a spring of spring constant k is given by

ν=12πkmν=12π×12003ν=3.183 Hz

Q. 14.9 (ii) A spring having with a spring constant 1200Nm1 is mounted on a horizontal table as shown in Fig. 14.24. A mass of 3kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0cm and released.

1424

Determine

(ii) maximum acceleration of the mass, and

Answer:

A body executing S.H.M experiences maximum acceleration at the extreme points

amax=FAmamax=kAmamax=1200×0.23amax=8ms2 (F A = Force experienced by body at displacement A from mean position)

Q. 14.9 (iii) A spring having with a spring constant 1200Nm1 is mounted on a horizontal table as shown in Fig. 14.24. A mass of 3kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0cm and released.

1424

Determine

(iii) the maximum speed of the mass.

Answer:

Maximum speed occurs at the mean position and is given by

vmax=Aωvmax=0.02×2π×3.18vmax=0.4ms1

Q. 14.10 (a) In Exercise 14.9, 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,

In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Answer:

Amplitude is A = 0.02 m

Time period is ω

ω=kmω=12003ω=20 rad/s

(a) At t = 0 the mass is at mean position i.e. at t = 0, x = 0

x(t)=0.02sin(20t)

Here x is in metres and t is in seconds.

Q. 14.10 (b) In Exercise 14.9, 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

(b) at the maximum stretched position,

In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Answer:

Amplitude is A = 0.02 m

Time period is ω

ω=kmω=12003ω=20 rad/s

(b) At t = 0 the mass is at the maximum stretched position.

x(0) = A

ϕ=π2

x(t)=0.02sin(20t+π2)x(t)=0.02cos(20t)

Here x is in metres and t is in seconds.

Q. 14.10 (c) In Exercise 14.9, 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

(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?

Answer:

Amplitude is A = 0.02 m

Time period is ω

ω=kmω=12003ω=20 rad/s

(c) At t = 0 the mass is at the maximum compressed position.

x(0) = -A

ϕ=3π2

x(t)=0.02sin(20t+3π2)x(t)=0.02cos(20t)

Here x is in metres and t is in seconds.

The above functions differ only in the initial phase and not in amplitude or frequency.

Q. 14.11 Figures 14.25 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.

1650450107919

Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.

Answer:

(a) Let the required function be x(t)=asin(±ωt+ϕ)

Amplitude = 3 cm = 0.03 m

T = 2 s

ω=2πTω=πrad s

Since initial position x(t) = 0, ϕ=0

As the sense of revolution is clock wise

x(t)=0.03sin(ωt)x(t)=0.03sin(πt)

Here x is in metres and t is in seconds.

(b)Let the required function be x(t)=asin(±ωt+ϕ)

Amplitude = 2 m

T = 4 s

ω=2πTω=π2rad s

Since initial position x(t) = -A, ϕ=3π2

As the sense of revolution is anti-clock wise

x(t)=2sin(ωt+3π2)x(t)=2cos(π2t)

Here x is in metres and t is in seconds.

Q. 14.12 (a) 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=2sin(3t+π/3)

Answer:

x=2sin(3t+π/3)x=2cos(3t+π3+π2)x=2cos(3t+5π6)

The initial position of the particle is x(0)

x(0)=2cos(0+5π6)x(0)=2cos(5π6)x(0)=3cm

The radius of the circle i.e. the amplitude is 2 cm

The angular speed of the rotating particle is ω=3rad s1

Initial phase is

ϕ=5π6ϕ=150o

The reference circle for the given simple Harmonic motion is

1650450164265

Q. 14.12 (b) 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).

(b) x=cos(π/6t)

Answer:

x(t)=cos(π6t)x(t)=cos(tπ6)

The initial position of the particle is x(0)

x(0)=cos(0π6)x(0)=cos(π6)x(0)=32cm

The radius of the circle i.e. the amplitude is 1 cm

The angular speed of the rotating particle is ω=1rad s1

Initial phase is

ϕ=π6ϕ=30o

The reference circle for the given simple Harmonic motion is

1650450237789

Oscillations Excercise:

Question:


Q. 14.12 (d) 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).

(d) x=2cosπt

Answer:

x(t)=2cos(πt)

The initial position of the particle is x(0)

x(0)=2cos(0)x(0)=2cm

The radius of the circle i.e. the amplitude is 2 cm

The angular speed of the rotating particle is ω=πrad s1

Initial phase is

ϕ=0o

The reference circle for the given simple Harmonic motion is

1650450280990

Q. 14.13 (a) Figure 14.30

(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 14.30 (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. 14.30

(b) is stretched by the same force F.

1650450344807

(a) What is the maximum extension of the spring in the two cases?

Answer:

(a) Let us assume the maximum extension produced in the spring is x.

At maximum extension

F=Kxx=Fk

(b) Let us assume the maximum extension produced in the spring is x. That is x/2 due to force towards left and x/2 due to force towards right

F=kx2+kx2x=Fk

Oscillations Excercise:

Question:

Q. 14.13 (b) Figure 14.26 (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 14.26 (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. 14.26 (b) is stretched by the same force F .

1650450400570

(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?

Answer:

(b).(a) In Fig, (a) we have

F=-kx

ma=-kx

a=kmx

ω2=kmT=2πωT=2πmk

(b) In fig (b) the two equal masses will be executing SHM about their centre of mass. The time period of the system would be equal to a single object of same mass m attached to a spring of half the length of the given spring (or undergoing half the extension of the given spring while applied with the same force)

Spring constant of such a spring would be 2k

F=-2kx

ma=-2kx

a=2kmxω2=2kmT=2πωT=2πm2kT=π2mk

Q. 14.14 The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0m If the piston moves with simple harmonic motion with an angular frequency of 200rad/min, what is its maximum speed ?

Answer:

Amplitude of SHM = 0.5 m

angular frequency is

ω=200 rad/minω=3.33 rad/s

If the equation of SHM is given by

x(t)=Asin(ωt+ϕ)

The velocity would be given by

v(t)=dx(t)dtv(t)=d(Asin(ωt+ϕ))dtv(t)=Aωcos(ωt+ϕ)

The maximum speed is therefore

vmax=Aωvmax=0.5×3.33vmax=1.67ms1

Q. 14.15 The acceleration due to gravity on the surface of moon is 1.7ms2 What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5s ? (g on the surface of earth is 9.8 m s–2)

Answer:

The time period of a simple pendulum of length l executing S.H.M is given by

T=2πlg

g e = 9.8 m s -2

g m = 1.7 m s -2

The time period of the pendulum on the surface of Earth is T e = 3.5 s

The time period of the pendulum on the surface of the moon is T m

TmTe=gegmTm=Te×gegmTm=3.5×9.81.7Tm=8.4s

Q. 14.16 (a) Answer the following questions :

(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle:

T=2πmk. A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?

Answer:

In case of spring, the spring constant is independent of the mass attached whereas in case of a pendulum k is proportional to m making k/m constant and thus the time period comes out to be independent of the mass of the body attached.

Q. 14.16 (b) Answer the following questions :

(b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations: For larger angles of oscillation, a more involved analysis shows that T is greater than 2πlg. Think of a qualitative argument to appreciate this result.

Answer:

In reaching the result T=2πlg we have assumed sin(x/l)=x/l. This assumption is only true for very small values of x (or θ) . Therefore it is obvious that once x takes larger values we will have deviations from the above-mentioned value.

Q. 14.16 (c) Answer the following questions :

(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?

Answer:

The watch must be using an electrical circuit or a spring system to tell the time and therefore free falling would not affect the time his watch predicts.

Q. 14.16 (d) Answer the following questions :

(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?

Answer:

While free falling the effective value of g inside the cabin will be zero and therefore the frequency of oscillation of a simple pendulum would be zero i.e. it would not vibrate at all because of the absence of a restoring force.

Q .14.17 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?

Answer:

Acceleration due to gravity = g (in downwards direction)

Centripetal acceleration due to the circular movement of the car = a c

ac=v2R (in the horizontal direction)

Effective acceleration is

g=g2+ac2g=g2+v4R2

The time period is T'

T=2πlgT=2πlg2+v4R2

Q. 14.18 A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρı . The cork is depressed slightly and then released

Show that the cork oscillates up and down simple harmonically with a period T=2πhρρ1g where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).

Answer:

Let the cork be displaced by a small distance x in downwards direction from its equilibrium position where it is floating.

The extra volume of fluid displaced by the cork is Ax

Taking the downwards direction as positive we have

ma=ρ1gAxρAha=ρ1gAxd2xdt2=ρ1gρhx

Comparing with a=-kx we have

k=ρ1gρhT=2πkT=2πρhρ1g

Q. 14.19 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.

Answer:

Let the height of each mercury column be h.

The total length of mercury in both the columns = 2h.

Let the cross-sectional area of the mercury column be A.

Let the density of mercury be ρ

When either of the mercury columns dips by a distance x, the total difference between the two columns becomes 2x.

Weight of this difference is 2Axρg

This weight drives the rest of the entire column to the original mean position.

Let the acceleration of the column be a Since the force is restoring

2hAρ(a)=2xAρga=ghx

d2xdt2=ghx which is the equation of a body executing S.H.M

The time period of the oscillation would be

T=2πhg

NCERT solutions for class 11 physics chapter 14 oscillations: Additional Exercise Solution

Q. 14.20 An air chamber of volume V has a neck area of cross section a into which a ball of mass m just fits and can move up and down without any friction (Fig.14.33). Show that when the ball is pressed down a little and released , it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal [see Fig. 14.33].

1650450479295

Answer:

Let the initial volume and pressure of the chamber be V and P.

Let the ball be pressed by a distance x.

This will change the volume by an amount ax.

Let the change in pressure be ΔP

Let the Bulk's modulus of air be K.

K=ΔPΔV/VΔP=KaxV

This pressure variation would try to restore the position of the ball.

Since force is restoring in nature displacement and acceleration due to the force would be in different directions.

F=aΔPmd2xdt2=aΔpd2xdt2=ka2mVx

The above is the equation of a body executing S.H.M.

The time period of the oscillation would be

T=2πamVk

Q. 14.21 (a) You are riding in an automobile of mass 3000kg. . Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50o/o during one complete oscillation. Estimate the values of

(a) the spring constant K

Answer:

Mass of automobile (m) = 3000 kg

There are a total of four springs.

Compression in each spring, x = 15 cm = 0.15 m

Let the spring constant of each spring be k

4kx=mgk=3000×9.84×0.15k=4.9×104 N

Q. 14.21 (b) You are riding in an automobile of mass 3000kg . Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50o/o during one complete oscillation. Estimate the values of

(b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750kg. .

Answer:

The amplitude of oscillation decreases by 50 % in one oscillation i.e. in one time period.

T=2πmkT=2π×30004×4.9×104T=0.77 s

For damping factor b we have

x=x0e(bt2m)

x=x 0 /2

t=0.77s

m=750 kg

e0.77b2×750=0.5ln(e0.77b2×750)=ln0.50.77b1500=ln2b=0.693×15000.77b=1350.2287 kg s1

14.22 Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

Answer:

Let the equation of oscillation be given by x=Asin(ωt)

Velocity would be given as

v=dxdtv=Aωcost(ωt)

Kinetic energy at an instant is given by

K(t)=12m(v(t))2K(t)=12m(Aωcos(ωt))2K(t)=12mA2ω2cos2ωt

Time Period is given by

T=2πω

The Average Kinetic Energy would be given as follows

Kav=0TK(t)dt0TdtKav=1T0TK(t)dtKav=1T0T12mA2ω2cos2ωt dtKav=mA2ω22T0Tcos2ωt dtKav=mA2ω22T0T(1+cos2ωt2)dt

Kav=mA2ω22T[t2+sin2ωt4ω]0TKav=mA2ω22T[(T2+sin2ωT4ω)(0+sin(0))]Kav=mA2ω22T×T2Kav=mA2ω24

The potential energy at an instant T is given by

U(t)=12kx2U(t)=12mω2(Asin(ωt))2U(t)=12mω2A2sin2ωt

The Average Potential Energy would be given by

Uav=0TU(t)dt0TdtUav=1T0T12mω2A2sin2ωt dtUav=mω2A22T0Tsin2ωt dtUav=mω2A22T0T(1cos2ωt)2dt

Uav=mω2A22T[t2sin2ωt4ω]0TUav=mω2A22T[(T2sin2ωT4ω)(0sin0)]Uav=mω2A22T×T2Uav=mω2A24

We can see K av = U av

Q .14.23 A circular disc of mass 10kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5s. The radius of the disc is 15cm . Determine the torsional spring constant of the wire. (Torsional spring constant a is defined by the relation J=aθ , where J is the restoring couple and θ the angle of twist).

Answer:

J=aθ

Moment of Inertia of the disc about the axis passing through its centre and perpendicular to it is

I=MR22

J=Id2θdt2aθ=MR22d2θdt2d2θdt2=2aMR2θ

The period of Torsional oscillations would be

T=2πMR22aa=2π2MR2T2a=2π2×10×(0.15)2(1.5)2a=1.97 N m rad1

Q. 14.24 (a) A body describes simple harmonic motion with an amplitude of 5cm and a period of 0.2s. Find the acceleration and velocity of the body when the displacement is

(a) 5cm

Answer:

A = 5 cm = 0.05 m

T = 0.2 s

ω=2πTω=2π0.2ω=10π rad s1

At displacement x acceleration is a=ω2x

At displacement x velocity is v=ωA2x2

(a)At displacement 5 cm

v=10π(0.05)2(0.05)2v=0

a=(10π)2×0.05a=49.35ms2

Q. 14.24 (b) A body describes simple harmonic motion with an amplitude of 5cm and a period of 0.2s. Find the acceleration and velocity of the body when the displacement is

(b) 3cm

Answer:

A = 5 cm = 0.05 m

T = 0.2 s

ω=2πTω=2π0.2ω=10π rad s1

At displacement x acceleration is a=ω2x

At displacement x velocity is v=ωA2x2

(a)At displacement 3 cm

v=10π(0.05)2(0.03)2v=10π0.0016v=10π×0.04v=1.257ms1

a=(10π)2×0.03a=29.61ms2

Q. 14.24 (c) A body describes simple harmonic motion with an amplitude of 5cm and a period of 0.2s. Find the acceleration and velocity of the body when the displacement is

(c) 0cm

Answer:

A = 5 cm = 0.05 m

T = 0.2 s

ω=2πTω=2π0.2ω=10π rad s1

At displacement x acceleration is a=ω2x

At displacement x velocity is v=ωA2x2

(a)At displacement 0 cm

v=10π(0.05)2(0)2v=10π×0.05v=1.57ms1

a=(10π)2×0a=0

Q. 14.25 A mass attached to a spring is free to oscillate, with angular velocity ω , in a horizontal plane without friction or damping. It is pulled to distance x0 and pushed towards the centre with a velocity v0 at time t=0 Determine the amplitude of the resulting oscillations in terms of the parameters ω , x0 and v0 . [Hint : Start with the equation x=acos(ωt+θ) and note that the initial velocity is negative.]

Answer:

At the maximum extension of spring, the entire energy of the system would be stored as the potential energy of the spring.

Let the amplitude be A

12kA2=12mv02+12kx02A=x02+mkv02

The angular frequency of a spring-mass system is always equal to km

Therefore

A=x02+v02ω2

oscillation class 11 physics

oscillations class 11 ncert solutions cover the theory and applications of the concept that any material medium can be represented as a collection of many interconnected oscillators, forming a medium of waves. The chapter explains this concept in detail.

When using the oscillations ncert solutions, students will gain an understanding of the various factors that impact the movement of an object. These factors include period, frequency, harmonic motion, and uniform circular motion.class 11 physics chapter 14 ncert solutions offers a comprehensive study of these important topics, including details such as systems that execute harmonic motion, like springs, forced oscillations, force laws, velocity, and more. The solutions have been prepared with careful attention to even the smallest details, in order to provide students with a thorough understanding of the subject matter.

NCERT solutions for class 11 physics chapter-wise

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Oscillations Class 11 NCERT Topics

14.1 Introduction

14.2 Periodic and oscillatory motions

14.3 Simple harmonic motion

14.4 Simple harmonic motion and uniform circular motion

14.5 Velocity and acceleration in simple harmonic motion

14.6 Force law for simple harmonic motion

14.7 Energy in simple harmonic motion

14.8 Some systems executing simple harmonic motion

14.9 Damped simple harmonic motion

14.10 Forced oscillations and resonance

Other types of oscillatory motions that are discussed in the Oscillations Class 11 Chapter 14 are damped oscillations and forced oscillations. In damped oscillation, as the name indicates the oscillation gets damped after an interval of time.

We have seen the oscillation of a pendulum, this oscillation will be damped unless an external force is applied to maintain the oscillation. Such maintained oscillation due to an external agency is called forced or driven oscillations.

Importance of NCERT Solutions for Class 11 Physics Chapter 14 Oscillations:

On an average 6.67 % of questions from oscillation and waves are asked for JEE Mains. Most of the previous JEE mains questions from oscillation asked are from topics SHM and simple pendulum. For NEET exam 2 questions are expected from oscillation. The CBSE NCERT solutions for Class 11 Physics chapter 14 Oscillations will help to score well in Class 11 and competitive exams.

Key features of NCERT Solutions for Class 11 Physics Chapter 14

  1. Comprehensive Coverage: These ch 14 physics class 11 ncert solutions cover all the important topics and questions presented in the chapter, ensuring a thorough understanding of oscillatory motion.

  2. Exercise and Additional Exercise Solutions: Detailed solutions for oscillation exercise class 12 questions (14.1 to 14.19) and additional exercise questions (14.20 to 14.25) are provided, facilitating practice and self-assessment.

  3. Clarity and Simplicity: The oscillation chapter class 11 solutions are explained in clear and simple language, making complex concepts more accessible to students.

  4. Problem-Solving Skills: By practising with these solutions, students develop strong problem-solving skills, which are essential in physics and other subjects.

  5. Exam Preparation: These solutions are designed to help students prepare effectively for their exams, including board exams and competitive exams like JEE and NEET.

NCERT solutions for class 11 Subject wise

Also Check NCERT Books and NCERT Syllabus here

Subject wise NCERT Exemplar solutions

Frequently Asked Questions (FAQs)

1. What is the weightage of Oscillations Class 11 for NEET

From the NCERT chapter oscillations, two questions can be expected for NEET exam. For more questions solve NEET previous year papers.

2. Is the chapter oscillation important for JEE Main

Yes, oscillation is important for JEE Main. One or two question can be expected from oscillations for JEE Main. It is one of the important chapter for scholarship exams like KVPY and NSEP. To solve more problems on Oscillations refer to NCERT book, NCERT exemplar and JEE main previous year papers.

Articles

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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