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NCERT Exemplar Class 11 Physics Solutions Chapter 14 Oscillations

NCERT Exemplar Class 11 Physics Solutions Chapter 14 Oscillations

Edited By Vishal kumar | Updated on Apr 09, 2025 12:49 PM IST

Do you ever wonder what makes a swinging pendulum or a vibrating guitar string oscillate in a rhythm? NCERT Exemplar Class 11 Physics Solutions Chapter 14 steps into the interesting realm of oscillations! From basic harmonic motion to energy transitions in a swinging motion, this chapter discusses concepts that design things like clocks and musical instruments. Discover solved exemplar problems and ace these thrilling topics with ease.

This Story also Contains
  1. NCERT Exemplar Class 11 Physics Solutions Chapter 14: MCQI
  2. NCERT Exemplar Class 11 Physics Solutions Chapter 14: MCQII
  3. NCERT Exemplar Class 11 Physics Solutions Chapter 14: Very Short Answer
  4. NCERT Exemplar Class 11 Physics Solutions Chapter 14: Short Answer
  5. NCERT Exemplar Class 11 Physics Solutions Chapter 14: Long Answer
  6. NCERT Exemplar Class 11 Physics Solutions Chapter 14 Oscillations Topics
  7. Important Concepts and Formulas - NCERT Exemplar Class 11 Physics Solutions Chapter 14
  8. What will students learn in NCERT Exemplar Class 11 Physics Solutions Chapter 14?
  9. NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
  10. NCERT Exemplar Class 11th Solutions
NCERT Exemplar Class 11 Physics Solutions Chapter 14 Oscillations
NCERT Exemplar Class 11 Physics Solutions Chapter 14 Oscillations

NCERT Exemplar Class 11 Physics Solutions Chapter 14 Oscillations is very important for exams. In this chapter, we will learn about different types of oscillations, their properties and formulas. Solving exemplar problems helps the students to build a strong foundation for competitive exams. For easy learning, students can also download the NCERT Exemplar Class 11 Physics Chapter 14 PDF for future reference.

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NCERT Exemplar Class 11 Physics Solutions Chapter 14: MCQI

Question:14.1

The displacement of a particle is represented by the equation y=3cos(π42ωt). The motion of the particle is
(a) simple harmonic with period 2pw
(b) simple harmonic with period πω
(c) periodic but not simple harmonic
(d) non-periodic

Answer:

The answer is option (b), Simple harmonic with period πω

Question:14.2

The displacement of a particle is represented by the equation y=sin3ωt. The motion is
(a) non-periodic
(b) periodic but not simple harmonic
(c) simple harmonic with period 2πω
(d) simple harmonic with period πω

Answer:

The answer is option (b), periodic but not simple harmonic

Question:14.3

The relation between acceleration and displacement of four particles are given below:
(a) ax=+2x
(b) ax=+2x2
(c) ax=2x2
(d) ax=2x
Which one of the particle is exempting simple harmonic motion?

Answer:

The answer is option (d) ax=2x
Explanation: In simple harmonic motion,
Acceleration proportional (as well as opposite) to displacement.
Thus, opt (d)

Question:14.4

The motion of an oscillating liquid column in a U-tube is
(a) periodic but not simple harmonic
(b) non-periodic
(c) simple harmonic and time period is independent of the density of the liquid
(d) simple harmonic and time period is directly proportional to the density of the liquid

Answer:

The answer is the option (c) Simple harmonic and time period is independent of the density of the liquid.
Explanation: Let us take a test tube viz., filled with a liquid of density ρ up to height ‘h’.
When the liquid is lifted in arm Q to a height ‘y’ from A to B, the liquid in arm P drops by the height ‘y’ from A’ to C’. The height difference between the two arms is 2y.
Here, the hydrostatic pressure provides the restoring force, thus,
F=V.ρ.g
=A.2yρg , where,
A = Area of a cross-section of tube, & Fy,
Thus, it is simple harmonic motion.
T=2πm(inertia)k(spring)=2πA(2h)ρ2Aρg
Thus, T=2πhg
Thus, the motion is harmonic as the time period is independent of density.

Question:14.5

A particle is acted simultaneously by mutually perpendicular simple harmonic motion x=acosωtandy=asinωt. The trajectory of motion of the particle will be
(a) an ellipse
(b) a parabola
(c) a circle
(d) a straight line

Answer:

The answer is the option (c) a circle.
Explanation: We know that,
Resultant displacement = x + y
=acosωt+asinωt
Thus, y=a(cosωt+asinωt)
=a2[cosωt2+sinωt2]
=a2[cosωtcos45+sinωtsin45]
……. (particle is acted simultaneously by mutually perpendicular direction)
y=a2coss(ωt45) ……. Thus, the displacement can neither be a straight line nor a parabola
Now, let us square and add x & y,
x2+y2=a2cos2ωt+a2sin2ωt
Thus, x2+y2=a2
This is the equation of a circle, hence, opt (c)

Question:14.6

The displacement of a particle varies with time according to the relation
y=asint+bcost
(a) The motion is oscillatory but not SHM
(b) The motion is SHM with amplitude a + b
(c) The motion is SHM with amplitude a2+b2
(d) The motion is SHM with amplitude a2+b2

Answer:

The answer is the option (d) The motion is SHM with amplitude a2+b2
Explanation: Given: y=asinωt+bcosωt
Now, we know that,
The amplitude of motion, A=a2+b2
Thus, opt (d).

Question:14.7

Four pendulums A, B, C and D are suspended from the same elastic support as shown in the figure. A and C are of the same length, while B is smaller than A and D is larger than A. If A is given a transverse displacement,
(a) D will vibrate with maximum amplitude
(b) C will vibrate with maximum amplitude
(c) B will vibrate with maximum amplitude
(d) All the four will oscillate with equal amplitude.

Four pendulums are suspended from the same elastic support

Answer:

The answer is the option (b) C will vibrate with maximum amplitude.
Explanation: If the pendulum vibrates with transverse vibration,
Time period, T=2πlg, where,
l = length of pendulum A & C.
Now, the disturbance produced is transmitted to all pendulums, i.e., B, C & D, where the time period (T) of C & A is the same. C will vibrate with a maximum in resonance, as the periodic force of period T produces resonance in C.
Hence, opt (b).

Question:14.8

Figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is

Circular motion of a particle
(a)x(t)=Bsin(2πt30)
(b)x(t)=Bsin(πt15)
(c)x(t)=Bsin(πt15+π2)
(d)x(t)=Bcos(πt15+π2)

Answer:

The answer is the option (a)x(t)=Bsin(2πt30)
Explanation: Circular motion is executed by P with radius B.
Here, x=OQcos(90θ)
=OQsinθ
= Bsinω
Thus, θ=ωt
x=Bsin2πTt
x=Bsin2π30t
Thus, opt (a)

Question:14.9

The equation of motion of a particle is x=acos(αt)2.
The motion is
(a) periodic but not oscillatory.
(b) periodic and oscillatory.
(c) oscillatory but not periodic.
(d) neither periodic nor oscillatory.

Answer:

The answer is the option (c) Oscillatory but not periodic.

Question:14.10

A particle executing S.H.M. has a maximum speed of 30cms and a maximum acceleration of 60cms2. The period of oscillation is
(a)πs.
(b)(π2)s.
(c)2πs.
(d)(πt)s.

Answer:

Answer: The answer is the option (a)πs
Explanation: Let us consider the equation of a SHM
Thus,y=asinωt
v=dydt
=aωcosωt ……. (i)
a=dvdt
=aω2sinωt ….. (ii)
Vmax=30cms …. (given)
From (i),
Vmax=aω
Thus, aω=30 ….. (iii)
From (ii),
Now, amax=aω2=60 …….. (iv)
Thus, 60=ω×30
Therefore, ω=2rads
Now, 2πT=2
Thus, T=π.
Hence opt (a).

Question:14.11

When a mass m is connected individually to two springs S1 and S2, the oscillation frequencies are v1 and v2. If the same mass is attached to the two springs, as shown in Figure, the oscillation frequency would be

Two Springs connected to the same mass

Answer:

Answer: The answer is the option (b)v12+v22
Explanation: If we connect a mass (m) to a spring on a frictionless horizontal surface, then, their frequencies will be –
V1=12πK1m & V2=12πK2m
…. (i) …. (ii)
Since the springs are parallel, their equivalent will be-
Kp=K1+K2
& frequency will be –
Vp=V1=12πK1+K2m
=12π[K1m+K2m]12
From (i),
K1m=(2πv1)2=4π2v12
& K2m=(2πv1)2=4π2v22
Thus, Vp=12π[4π2v12+4π2v22]
Vp=v12+v22
Hence, opt (b)

NCERT Exemplar Class 11 Physics Solutions Chapter 14: MCQII

Question:14.12

The rotation of earth about its axis is
(a) periodic motion
(b) simple harmonic motion
(c) periodic but not simple harmonic motion
(d) non-periodic motion

Answer:

The answer is the option (a) periodic motion and (c) periodic but not SHM
Explanation:
(i) The earth completes one revolution in a regular interval of time.
(ii) The motion of the earth is circular about its own axis.
(iii) But this motion is not SHM as we cannot measure its displacement since it is not about a fixed point. Also, it does not move both sides.

Question:14.13

Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point is
(a) simple harmonic motion
(b) non-periodic motion
(c) periodic motion
(d) periodic but not SHM

Answer:

The answer is the option (a) Simple harmonic motion and (b) non - periodic motion
Explanation: Lift the ball from point A to B and smoothly release it to reach C & then return to first A and then B. Thus, it is periodic motion. Here, mg sin ? balances the restoring force (R), also, a restoring force (mgsinθ)acts on the ball.
Thus, ma=mgsinθ
a=gsinθ or d2xdt2=gsinθ ….. (when the ball moves upward)
d2xdt2=g(xr)
Thus, d2xdt2(x)
Thus, it is a simple harmonic motion.
Ω=gr or T=2πω=2πrg
Thus, this motion is periodic as well as simple harmonic.

Question:14.14

Displacement versus time curve for a particle executing SHM is shown in figure. Choose the correct statements.
(a) Phase of the oscillator is same at t = 0s and t = 2 s
(b) Phase of the oscillator is same at t = 2s and t = 6 s
(c) Phase of the oscillator is same at t = 1s and t = 7 s
(d) Phase of the oscillator is same at t = 1s and t = 5 s

Displacement - time graph in SHM

Answer:

The answer is the option (b) Phase of the oscillator is same at t = 2s & t = 6s and (d) The phase of the oscillator is the same at t = 1s & t = 5s
Explanation: If the mode of vibration of two particles is the same, then they are said to be in the same phase. That means, their distance will be nλ, where, n = 1, 2, 3,4, …..
(a) Here the particles are not in the same phase as the distance between them is λ2.
(b) The particles are at a distance λ. Thus, they are in the same phase.
(c) The distance between particles is λ=λ2=3λ2. Thus, they are not in the same phase.
(d) The distance between the particles is λ. Thus, they are in the same phase.

Question:14.15

Which of the following statements is/are true for a simple hannonic oscillator?
(a) Force acting is directly proportional to displacement from the mean position and opposite to it
(b) Motion is periodic
(c) Acceleration of the oscillator is constant
(d) The velocity is periodic

Answer:

The answer is the option (a) Force acting is directly proportional to the displacement from the mean position and opposite to it, (b) Motion is periodic, and (d) The velocity is periodic.
Explanation: Let, x=asinωt, be a SHM …… (i)
V=dxdt=αωcosω …. (ii)
V=dvdt=αω(ωsinωt)
Thus, A=αω2sinωt or A = -ω2x
mA=mω2x
Hence, F(x)
Now, x(t) = x(t + T)
Thus, the motion is periodic and simple harmonic.
Now, from (ii),
v(t) = v(t + T)
Hence, opt, (a), (b) & (d) are correct options.

Question:14.16

The displacement time graph of a particle executing S.H.M. is shown in Figure. Which of the following statement is/are true?
Displacement - time graph in SHM

(a) The force is zero at t=(3T4)
(b) The acceleration is maximum at t=(4T4)
(c) The velocity is maximum at t=(T4)
(d) The P.E. is equal to K.E. of oscillation at t=(T2)

Answer:

The answer is the option (a) The force is zero at t=(3T4), (b) The acceleration is maximum at t=(4T4), and (c) The velocity is maximum at t=(T4)
Explanation: (a) The particle is at its mean position at t=(3T4), so the force acting on it zero, but due to inertia of mass the motion continues
a = 0 thus, F = 0.
(b) Particles velocity changes increasing to decrease so maximum at the change in velocity, at t=(4T4). Thus, acceleration is maximum here.
(c) The velocity is maximum at its mean position at t=(T4) as there is no retarding force on it.
(d) K.E. = 0 at t=T2=2T4.
Thus, P.E is not equal to kinetic energy.

Question:14.17

A body is performing SHM, then its
(a) The average total energy per cycle is equal to its maximum kinetic energy
(b) The average kinetic energy per cycle is equal to half of its maximum kinetic energy
(c) mean velocity over a complete cycle is equal to 2π times of its maximum velocity
(d) root mean square velocity is 12 times of its maximum velocity

Answer:

The answer is the option (a) Average total energy per cycle is equal to its maximum kinetic energy, (b) Average kinetic energy per cycle is equal to half its maximum kinetic energy, and (d) Root square mean velocity is equal to 12 times its maximum velocity.
Explanation: (a) letx=asinωt be a periodic SHM
Let m be the mass executing SHM
v=dxdt
= aωcosωt
vmax=αω.....(sincecosωt=1)
Now, K.E.P.E = Total mechanical energy
Thus, T.E.=12ma2ω2
Thus, K.E.=12ma2ω2
Or we can say that the average total energy is K.E.max
(b) Let, amplitude = a
Angular frequency = ω
Thus, maximum velocity = αω, which varies according to the sine law.
Thus, the rms value of a complete cycle = 12αω.
Thus, the average
K.E.=12mvrms2
= 12m12a2ω2
= 12[m12a2ω2]
= 12[12mmax2]
=12K.E.max
(c) v=aωcosωt
Vmean=(aωaω)2
Vmean = 0 ….. since vmax is not equal to vmean
Vrms=v12+v222=aω2
Thus, vrms=vmax2

Question:14.18

A particle is in linear simple harmonic motion between two points A and B, 10 cm apart (figure). Take the direction from A to B as the positive direction and choose the correct statements. __ _
AO = OB = 5 cm
BC= 8 cm
Particle in Linear SHM
(a) The sign of velocity, acceleration and force on the particle when it is 3 cm away from A going towards B are positive
(b) The sign of velocity of the particle at C going towards B is negative
(c) The sign of velocity, acceleration and force on the particle when it is 4 cm away from B going towards A are negative
(d) The sign of acceleration and force on the particle when it is at point B is negative

Answer:

The answer is the option (a) The sign of velocity, acceleration & force on the particle when it is 3 cm away from A going towards B is positive, (c) The sign of velocity, acceleration & force on the particle when it is 4 cm away from B going towards A are negative, and (d) The sign of acceleration and force on the particle when it is at point B is negative.
Explanation: (a) The velocity of the particle increases up to 0 when it is 3 cm away from A and is going from A to B, i.e., in the positive direction. Thus, the velocity is positive. Also, acceleration in SHM is towards positive.
(b) Here the velocity is positive and not negative since the particle is going towards B.
(c) Here the particle is 4 cm away from B and is going towards A, i.e., the particle is going from B to A, i.e., in the negative direction. Hence, the velocity & acceleration towards mean position O is negative.
(d) Here, force and acceleration both are negative as the particle is at B, and they both are towards O.

NCERT Exemplar Class 11 Physics Solutions Chapter 14: Very Short Answer

Question:14.19

Displacement versus time curve for a particle executing SHM is shown in figure. Identify the points marked at which (i) velocity of the oscillator is zero, (ii) speed of the oscillator is maximum.

Answer:

(i) At A, C, E, and G the displacement is maximum. Hence, the velocity of the oscillator will also be maximum.
(ii) At B, D, F and H, the displacement of the oscillator is zero. Thus, there is no restoring force. Hence, the speed of the oscillator will be maximum.

Question:14.20

Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shown in figure. When the mass is displaced from equilibrium position by a distance x towards right, find the restoring force.

Two springs connected to the same block

Answer:

If we displace the mass ‘m’ frpm its equilibrium position towards the right by distance ‘x’, The spring B will be compressed by distance x, & let kx be the force applied on the mass ‘m’ towards left. If we apply the force kx on the mass, A will be extended by distance x towards left; and apply force kx towards left. Thus, restoring force will act on the block as net force towards left.
F = kx + kx
= 2kx
Thus, restoring force towards the left is 2kx.

Question:14.21

What are the two basic characteristics of a simple harmonic motion?

Answer:

(i)The direction of acceleration is towards the mean position, and
AccelerationDisplacement (from mean position)
(ii) The direction of force and displacement are opposite, thus, F=kx.
RestoringforceDisplacement.
These are the two basic characteristics of SHM.

Question:14.22

When will the motion of a simple pendulum be simple harmonic?

Answer:

Let us consider a pendulum whose,
Length = l
Mass of bob = m, viz., displaced by angle = θ
Now, Restoring force =F=mgsinθ
Also, if θ is small then
Sin θ = θ
=arcradius
=xl
Thus,
F=mgxl
OrF(x) …………. (since m, g & l are constants)
Thus, simple harmonic for the small-angle

Question:14.21

What is the ratio of maximum acceleration to the maximum velocity of a simple harmonic oscillator?

Answer:

Let us consider x=Asinωt to be an SHM
Now, v=dxdt
=Aωcosωt
Now, for vmax,cosωt=1
Thus, vmax=Aω
Now, a=dvdt
=Aω2sinωt
Now, for amax,
sinωt=1
Thus, amax=Aω2
Now, amaxvmax=Aω2Aω
=ω.

Question:14.24

What is the ratio between the distance travelled by the oscillator in one time period and amplitude?

Answer:

We already know that the distance travelled by an oscillator in a one-time period is equal to 4A,
Where A = amplitude of oscillation.
Thus, we know that the
Required ratio
4AA
=4:1

Question:14.25

In figure, what will be the sign of the velocity of the point P’, which is the projection of the velocity of the reference particle P. P is moving in a circle of radius R in anti-clockwise direction
Circular motion of a particle

Answer:

Circular motion of a particle
At time = t, P’ is the foot perpendicular of the velocity vector of particle P.
This foot shifts from P to Q, i.e., towards the negative axis, when the particle moves from P to P1.
Thus, there is –ve sign of θ motion of P’.

Question:14.26

Show that for a particle executing SHM, velocity and displacement have a phase difference of π2.

Answer:

Let us consider x=Asinωt to be a SHM …….. (i)
Now, v=dxdt
=Aωcosωt
=Aωsin(90+ωt) …….. [since sin(90+θ)=cosθ]
Thus, v=Aωsin(ωt+(π2))
From (i), we know that,
The phase of displacement = ωt
& from (ii), we know that,
The phase of velocity = (ωt+(π2))
Thus, the phase difference = ωt+π2ωt
=π2.

Question:14.27

Draw a graph to show the variation of PE, KE and total energy of a simple harmonic oscillator with displacement.

Answer:

Energy Time graph in SHM
Let us consider that a mass is lying on a horizontal frictionless surface, where spring constant = k.
If we displace the mass by distance A from its mean position, then it will execute SHM.
At this stretched position, P.E. of mass = 12kA2.
Now, at maximum stretch, i.e.,
At x = A, K.E. = 0
Now, at x=x±A
P.E. = total energy
= 12kA2.
Now let’s consider that the mass is back at its mean position, now the restoring force acting on the particle will be
P.E.=12kx2
= 12mω2x2
The restoring force constant of oscillator isk=mω2
When x±A
P.E.=12mω2A2
& K.E.=12mv2
=12mω2(A2x2)
Now, when x=±A,K.E.=0
xK.E.P.E.T.E.
012mω2A2012mω2A2
+A012mω2A212mω2A2
-A012mω2A212mω2A2
Now, at displacement ‘x’, T.E. will be,
E = K.E. + P.E.
= 12mω2A2
Thus, with displacement ‘x’, E is constant.

Question:14.28

The length of a second’s pendulum on the surface of earth is 1 m. What will be the length of a second’s pendulum on the moon?

Answer:

A second pendulum is a pendulum with time period (T) = 2 seconds
Te=2πlege
Thus, Te2=4π2lege …… (i)
Now, Tm=2πlmgm
Thus, Tm2=4π26lmge ……. (ii) (gm=ge/6)
Now, for the second pendulum,
Tm2Te2=6gelegelm
1=6lmle
1=6lm1m ……. (since le=1m)
Thus, lm=16m

NCERT Exemplar Class 11 Physics Solutions Chapter 14: Short Answer

Question:14.29

Find the time period of mass M when displaced from its equilibrium position and then released for the system as shown in figure.
Block - pulley system in shm

Answer:

If we pull mass M & then release it, it oscillates with the pulley up & down. Let x0 be the extension of the string when loaded with M, Due to acceleration and the same amount of forces the extension and compression of the spring from initial position is larger and smaller, respectively. Hence, we can neglect the gravitational force here.
Now let us apply force ‘F’ to pull M downwards by displacement x. As the string cannot be extended, its extension will be 2x.
Thus, the total extension = (x0+2x)
When we pull it downward by x
T=k(x0+2x)
When we do not pull M,
T=kx0
F = 2T
Thus, F=2kx0
& F=2T
F=2k(x0+2x)
Now, restoring force,
Frest =(FF)
= [2k(x0+2x)2kx0]
= 2k.2x
Thus, Ma=4kx
a=4kMx
Thus, ax
Therefore, it is a simple harmonic motion.
Now, a=ω2x
ω2=ax
=+4kM
ω=2kM
T=πMk

Question:14.3

Show that the motion of a particle represented byy=sinwtcoswt is simple harmonic with a period of 2πω.

Answer:


y=sinωtcosωt
=2 [sinωt12cosωt12]
=2 sin[ωtπ4]
Comparing this equation with standard SHM we get,
ω=2πT
Or t=2πω

Question:14.31

Find the displacement of a simple harmonic oscillator at which its PE is half of the maximum energy of the oscillator.

Answer:

Let us consider an oscillator viz. at ,
displacement = x from its mean position,
Mass = m
P.E.=12kx2
Force constant of oscillator, k=mω2
Thus, P.E.=m2x2
At x = A, when K.E. = 0, P.E. will be maximum & it will be the total energy of the oscillator
E=12mω2A2
At displacement ‘x’, P.E.=12E
Thus,
12mω2x2= 12.12mω2A2
& x=±A2
Thus, when displacement = ±12 amplitude from mean position, P.E. will be half of the total energy.

Question:13.32

A body of mass m is situated in a potential field U(x)=U0(1cosαx) when U0 and α are constants. Find the time period of small oscillations.

Answer:

Here, dW = F.dx
So, if W = U, dU = F.dx
Or F=dUxdx ….. (since, restoring force here is opposite to displacement)
F=ddx[U0(1cosαx)=ddx[U0+U0cosαx]
F=αU0sinαx
Now, αx is small for SHM, So sin αx will become αx …… (i)
Thus, F=αU0αx
Now, since, U0 & α are constants,
Fx
Thus, the motion will be SHM.
From (ii)
k=α2U0
mω2=α2U0
Thus, ω2=α2.U0m
(2πT)2=α2.U0m
T=2παmU0
Thus, considering (i) time period is valid for the small-angle αx.

Question:14.34

Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of 2 to the right with the vertical, the other pendulum makes an angle of 1 to the left of the vertical. What is the phase difference between the pendulums?

Answer:

Now, Θ1=Θ0sin(ωt+δ1)
& Θ2=Θ0sin(ωt+δ2)
Now, for the first pendulum,
Θ=2
Thus, sin(ωt+δ1)=1
& for second pendulum,
Θ=1
Thus, sin(ωt+δ2)=1
Thus, (ωt+δ1)=90 & (ωt+δ2)=30
Therefore,
δ1+δ2=120

NCERT Exemplar Class 11 Physics Solutions Chapter 14: Long Answer

Question:14.35

A person normally weighing 50 kg stands on a massless platform which oscillates up and down harmonically at a frequency of 2.0 s-1 and an amplitude 5.0 cm. A weighing machine on the platform gives the persons weight against time.
(a) Will there be any change in weight of the body, during the oscillation?
(b) If answer to part (a) is yes, what will be the maximum and minimum reading in the machine and at which position?

Answer:

Due to normal reaction ‘N’, there will be weight in weight machine.
Let us consider the top positions of the platform, both the forces, due to the weight of the person and oscillator acts downwards.
Thus, the motion will be downwards.
Let us consider, acceleration = a
ma = mg – N …….. (i)
Now, when the platform moves upwards from its lowest position,
ma = N – mg …….. (ii)
Now, acceleration of oscillator is
a=ω2A
From (i),
N=mgmω2A
Where,
Amplitude = A
Angular frequency = ω &
Mass of oscillator = m
ω=2πv
=4πrad/s
A = 5cm
=5×102m
m = 50 kg
N=50×9.850×4π×4π×5×102
=50[9.816π2×5×102]
=50×1.91
=95.5N
Thus, the minimum weight is 95.5 N.
From (ii),
N – mg = ma
Now, for upward motion from the lowest point of the oscillator,
N=m(a+g)
=m[9.81+ω2A]=ω2A
=50[9.81+7.89]
=50[17.7]
=885N
Therefore, during oscillation, there is a change in the weight of the body.
Also, maximum weight = 885 N, when the platform moves to upward direction from lowest direction & minimum weight = 95.5 N, when the platform moves to downward direction from the highest point.

Question:14.36

A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4cm below the point, where it was held in hand.
(a) What is the amplitude of oscillation?
(b) Find the frequency of oscillation?

Answer:

(a) As no deforming force acts on the spring when mass ‘m’ is supported by hand extension in the spring. Let m reach its new position at displacement = x unit, from the previous one, then,
P.E. of the spring or mass = gravitational P.E. lost by man
P.E. = mgx
But due to spring,
12kx2k=ω2A
Thus, x=2mgk
When extension is x0, the spring will be at the mean position by the block
F=kx0
F = mg
Thus, x0=mgk ……. (ii)
Now, from (i) & (ii)
x=2(mgk)
=2x0=4cm
Thus, x0=2cm
xx0=42=2cm.
Thus, from the mean position, the amplitude of the oscillator is maximum.
(b) Now, we know that time period (T) does not depend on amplitude,
T=2πmk
From (i)
2mgk=x
mk=x2g
=4×1022×9.8
Or, km=2×9.84×102
Now, v=12πkm
=12×3.142×9.84×102
=10×2.216.28
=3.52Hz
Since the total extension in the spring is 4 cm when released & amplitude is 2 cm, the oscillator will not rise above 4cm, & hence, it will oscillate below the released position.

Question:14.37

A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period.
T=2πmAρg
where m is mass of the body and ρ is density of the liquid.

Answer:

Cylindrical Log in SHM
If we press a log downward into the liquid, a buoyant force acts on it, and due to inertia it moves upwards from its mean position & comes down again due to gravity.
Thus, the restoring force on the block = Buoyant force (B.F.) – mg
Volume of liquid displaces by block = V
When it floats,
mg = BF
Or, mg=Vρg
mg=Ax0ρg …….. (i)
Area of crossection = A
Height of liquid block =x0
When pressed in water, the total height of the block in water = (x+x0)
Thus, net restoring force = [A(x+x0)]ρgmg
Frest =Axρg …….. ( since BF is upward & x is downward)
Frest proportional to-x
Hence, the motion here is SHM.
Now, kAρg
a=ω2xω
=km
Thus, T=2πmk
Frest =Aρgx
ma=Aρgx
a=Aρgxm
ω2x=Aρgxm
K=Aρg
(2π2T)=Aρgm
Thus,
T2π=mAρg
Thus,
T=2πmAρg

Question:14.38

One end of a V-tube containing mercury is connected to a suction pump and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of 45 each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.

Answer:

Let initial height = h0 of liquid in both columns; due to pressure difference, if the liquid in arm A is pressed by x, then liquid in arm B will rise by x.
Let us consider dx as an element of height, then,
Mass dm=A.dx.ρ,where, Area of a cross-section of tube = A
PE of left dm element columns = (dm) gh
& PE of dm in left columns = dm=A.dx.ρgx
Thus, total PE in left column = 0hAρgxdx=Aρg[x22]0h0
=Aρgh122
From the figure
sin45=h1l
h1=h2=lsin45=12
h12=h22=l22
Thus, PE in left column = PE in right column = Aρgl24
Thus, total potential energy = Aρgl22
Let the element move towards right by y units due to the pressure difference, then,
Liquid column in left arm = (l – y)
Liquid column in right arm = (l + y)
PE in left arm =Aρg(ly)2sin245
PE in right arm = Aρg(l+y)2sin245
Total PE=Aρg(12)2[(ly)2+(l+y)2]
Final PE=Aρg2[l2+y22ly+l2+y2+2ly]
=Aρg2(2l2+2y2)
Change PE = Final PE – Initial PE
=Aρg2(2l2+2y2)Aρgl22
=Aρg2(2l2+2y2)
If there is change in velocity(v) of liquid column,
ΔKE=12mv2
m=(A.2l)ρ
Thus, ΔKE=Aρlv2
Now, change in total energy = Aρg2(l2+2y2)+Aρlv2
Total change in energy,
ΔPE+ΔKE=0
Thus,Aρg2(l2+2y2)+Aρlv2=0
Aρ2[g(l2+2y2)+2lv2]=0
Aρ2 is not equal to 0, thus,
[g(l2+2y2)+2lv2]=0
Now, let us differentiate w.r.t. t,
g[0+2×2×2dydt]+2l.2v.dvdt=0
d2ydt2+gl.y=0
Since 4v is not equal to 0,
d2ydt2+ω2y=0
Thus, ω2=gl
2πT=gl
Thus,
T=2πlg

Question:14.39

A tunnel is dug through the centre of the Earth. Show that a body of mass ‘m’ when dropped from rest from one end of the tunnel will execute simple harmonic motion.

Answer:

We know that,
Acceleration due to gravity inside the earth = g’
g=g(1dR)
= =g(RdR)
Now, R-d = y
Thus, g=g.yR
On both of them, force at depth ‘d’ will be –
F=mg
mg.yR
F(y)
Thus, in the tunnel, the motion of the body is SHM.
We can say,
ma=mg
a=gRy
ω2y=gRy
Thus, 2πT=gR
Or, T=2πRg.

Question:14.40

A simple pendulum of time period 1s and length l is hung from a fixed support at O, such that the bob is at a distance H vertically above A on the ground (Figure). The amplitude is θ0. The string snaps at θ=θ02. Find the time taken by the bob to hit the ground. Also, find the distance from A where bob hits the ground. Assume θ0 to be small so that sinθ0θ0 and cos θ01.

Simple pendulum

Answer:

Let us consider the diagram at which,
t = 0
Θ=Θ02,Θ=Θ0cosωt
Now, at t = t1,
Θ=Θ02
Simple pendulum
Thus, Θ02=Θ0cos2πTt1
…….. (Given: T = 1sec)
Thus, cos2πt1=12=cosπ3
2πt1=π3
Or, t1=16
Θ=Θ0cos2πt
dΘdt=Θ02πsin2πt

Now, at t=16
Θ=Θ02
dΘdt=Θ02πsin2π16
=Θ02π32
=Θ0π3
Now, ω=Θ0π3
i.e., vl=Θ0π3
Thus, v=3Θ0πl
By the –ve sign it is clear that the bob’s motion is towards left.
Now,
vx=vcosΘ02
==3Θ0πlcosΘ02
vy=vsinΘ02=3πΘ0lsinΘ02
Let H’ be the vertical distance covered by vy,
s=ut+12gt2
H=vyt=12gt2
12gt2+3πΘ0lsinΘ02×tH=0
Thus,12gt2+3πΘ0lΘ02×tH=0 ……. (Given: sinΘ02=Θ02)
Now, by quadratic formula,
t=B±B24AC2A
t=3πθ02l4±(3π2θ04l2)24.12.gH2g2
We will neglect θ04 and θ02 since θ0 is very small.
Thus, t=+2gHgH=H+H"
Thus, 2Hg
H’’<<H’ as Θ02 is very small.
Thus, H = H’
Thus, vxt = distance covered in horizontal direction
X=3πθ0l.2Hg
(Given: cosθ02=1)
Thus, X=Θ0lπ6Hg
The bob was at a distance of lsinθ02=lθ02 from A at the time of snapping.
Thus, the distance of bob from A where it meets the ground is
=lΘ02X
=lΘ02Θ0lπ6Hg
=lΘ0[12π6Hg]

NCERT Exemplar Class 11 Physics Solutions Chapter 14 Oscillations Topics

Important Concepts and Formulas - NCERT Exemplar Class 11 Physics Solutions Chapter 14

1. Periodic and Oscillatory Motions
- Periodic Motion: Motion that repeats after a fixed interval of time.
- Oscillatory Motion: A type of periodic motion in which a body moves back and forth about a mean position.
2. Simple Harmonic Motion (SHM)

  • SHM is a special type of oscillatory motion where restoring force is directly proportional to displacement and acts towards the mean position.
  • Equation of SHM:
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F=kx

where k is the force constant and x is the displacement.
3. SHM and Uniform Circular Motion

  • SHM can be considered as the projection of uniform circular motion on the diameter of a circle.
  • Displacement equation in SHM:

x=Acos(ωt+ϕ)

where A is amplitude, ω is angular frequency, and ϕ is the phase constant.
4. Velocity and Acceleration in SHM

  • Velocity in SHM:

v=±wA2x2

  • Acceleration in SHM:

a=ω2x

where ω=k/m is the angular frequency.

5. Energy in SHM

  • Kinetic Energy (KE):

KE=12mω2(A2x2)

  • Potential Energy (PE):

PE=12mω2x2

  • Total Energy (TE):

TE=12mω2A2

Energy remains constant, exchanging between kinetic and potential forms.
6. Damped Simple Harmonic Motion

  • Damping: The gradual loss of energy in oscillations due to resistive forces like friction or air resistance.
  • Equation of Damped SHM:

md2xdt2+bdxdt+kx=0

where b is the damping constant.
7. Forced Oscillations and Resonance

  • Forced Oscillation: When an external periodic force is applied to an oscillating system.
  • Resonance: A condition where the frequency of external force matches the natural frequency, causing maximum amplitude.
  • Resonance Condition:

fexternal =fnaturnal 

What will students learn in NCERT Exemplar Class 11 Physics Solutions Chapter 14?

NCERT Exemplar Class 11 Physics Chapter 14 solutions covers wave motion, equations, and vibrations in open and closed pipes with practical applications. The solutions help in solving tricky questions and gaining a deeper grasp of the concepts.

NCERT Exemplar Class 11 Physics Solutions Chapter-Wise

NCERT Exemplar Class 11th Solutions

Check Class 11 Physics Chapter-wise Solutions

Also, Read NCERT Solution subject-wise -

Check NCERT Notes subject-wise -

Also, Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. Where do we find the answers to exemplar exercise questions of Class 11 Physics chapter oscillations?

 In this, NCERT Exemplar Class 11 Physics solutions Chapter 14 Oscillations, students can find the answers of all the exemplar questions of chapter 14. 

2. What kind of questions are asked in examinations regarding Class 11 Physics topic?

The question can be of MCQ, short answer or long answer type. All these types of questions are given after each chapter, the students need to go through them once before appearing for the examination.

3. What all topics are covered in NCERT Exemplar Class 11 Physics solutions Chapter 14?

Topics such as oscillations, it’s types, SHM and its characteristic features are covered in it. 

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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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