Class 11th Chapter -14 Oscillations |Physics| NCERT Solution | Edugrown

NCERT Exercises

Question 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 back.
(b) A freely suspended bar magnet displaced from its N-S direction and released.
(c) A hydrogen molecule rotating about its center of mass.
(d) An arrow released from a bow.
Answer:
(a) It is not a periodic motion as the swimmer completes only one trip but if he makes more than one trip and time for each trip is same, the motion can be categorised as periodic.
(b) It is a periodic motion because a freely suspended bar magnet if once displaced from N-S direction and released, it oscillates about this position. Hence it is simple harmonic motion also.
(c) It is also a periodic motion.
(d) It is not a periodic motion.

Question 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) It is periodic but not simple harmonic motion because it is not to and fro motion about a fixed point.
(b) It is simple harmonic motion.
(c) It is simple harmonic motion.
(d) It is periodic but not simple harmonic motion. A polyatomic molecule has a number of natural frequencies and its general motion is the resultant of simple harmonic motions of a number of different frequencies. The resultant motion is periodic but not simple harmonic motion.

Question 3.
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)?

Answer:
(a) Does not represent periodic motion, as the motion neither repeats nor comes to mean position.
(b) Represents periodic motion with period equal to 2 s.
(c) Does not represent periodic motion, because it is not identically repeated.
(d) Represents periodic motion with period equal to 2 s.

Question 4.
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 ωt-cos ωt
(b) Sin³ ωt
(c) 3 cos( -2ωt)
(d) Cos ωt +Cos 3ωt+cos 5ωt
(e) exp(-ω² t²)
(f) 1+ωt+ω²t²

Answer:
The function will represent a periodic motion, if it is identically repeated after a fixed interval of time and will represent simple harmonic motion, if it can be written

(d)    cos ωt+ cos 3ωt +cos 5 ωt
It represents periodic but not simple harmonic motion. Its time period is
 .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, cos ωt has a period T = , cos3 ωt has a period , cos5 ωt has period , the last two forms repeat after any integral multiple of their period. Thus each term in the sum repeats itself after T, and hence the sum is a periodic function with a period .
(e) exp (-ω² t²) : It is an exponential function which decreases monotonically with increasing time and tends to zero as t →∞ and thus never repeats itself. Therefore it represents non-periodic motion.
(f)  1 + ωt+ (-ω² t² ): It represents non-periodic motion (physically unacceptable because the function tends to infinity as t →∞).

Question 5.
A particle is in linear simple harmonic motion between two points, A and 6 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.
Answer:

Refer figure, here A and B represent the two extreme positions of a simple harmonic motion. For velocity, the direction from A to B is taken to be positive. For acceleration and the force, the direction is taken positive if directed along AP and negative if directed along BP.
(a) Harmonic motion is momentarily at rest being its extreme position of motion. Therefore, its velocity is zero, acceleration is +ve because it is directed along AP. Force is also +ve, since the force is directed along AP, i.e., +ve direction.
∴     0, + ,+
(b) At the end B, velocity is zero. Here, acceleration and force are negative as they are directed along BP, i.e., along negative direction.
∴   0, -, –
(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, i.e., -ve direction. Hence, velocity is -ve. Both, acceleration and force are zero.
∴   -, 0,0
(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.Here 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 B going towards A, the ,particle is at S with a tendency to move along SP which is negative direction,Here, velocity, acceleration and force, all are negative.
∴   -, -, –

Question 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) o = -200x²
(c) a = -10 x
(d) o=100x³
Answer:
A particle is said to be executing simple harmonic motion, if the acceleration ‘a’ produced in it satisfies the following two conditions:
(1) ‘a’ is directly proportional to the displacement (say y) from the mean position, i.e. a ∝ y.
(2) ‘a’ is directed towards mean position i.e. acts opposite to the direction in which y increases.
Mathematically a = -ω²y …(i)
where ω = angular frequency.

(a) a =7x does not satisfy eqn. (i), so it does not represent simple harmonic motion.
(b) a = -200.x², does not satisfy equation (i) hence it does not represent simple harmonic motion.
(c) a = -10x, here x = It satisfies equation (i), so it represents simple harmonic motion.
(d) a = 100x³ does not satisfy equation (i), hence it does not represent simple harmonic motion.

Question 7.
The motion of a particle executing simple harmonic motion is described by the displacement function.
x(t) = A cos(ωt+ φ)
If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm s^-1, what are its amplitude and initial phase angle? The angular frequency of the particle is Π s^-1 If instead of the cosine function, we choose the sine function to describe the simple harmonic motion : x = B sin ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.
Answer:
Here, at t = 0, x = 1 cm and υ= co cm s^_1;
ω= Π S^_1
Given : x = A cos(ωt + φ) … (i)
Since at t = 0, x = 1, we get
1 = A cos(Π x 0 +φ) = A cosφ
The instantaneous particle velocity is given by


Question 8.
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?
Answer:
Here, m = 50 kg,
Maximum extension, y = 20 – 0 = 20 cm = 0.2 m
Maximum force, F=mg = 50 x 9.8 = 490 N

Question 9.
A spring having 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.

Determine
(i) the frequency of oscillations of the mass,
(ii)maximum acceleration of the mass,and
(iii) the maximum speed of the mass.
Answer:
Here, m = 3.0 kg; k = 1200 N m^_1,
A = 2 cm = 0.02 m
(i) The frequency of oscillations of the attached mass is

Question 10.
In previous question let us take the position of mass when the spring is unstretched 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 stretched position, and
(c) at the maximum compressed position.
In what way do these functions for simple harmonic motion differ from each other, in frequency, in amplitude or the initial phase?
Answer:
From previous solution

(a) As time is noted from the mean position, hence using
x = A sin ωt, we get x = 2 sin 20t
(b) At maximum stretched position, the mass is at the extreme right position, with an  initial phase of  rad. Then,

Question 11.
Figures (a) and (b) 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.

Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
Answer:
In figure
(a) T = 2 s; A = 3 cm
At t = 0, OP makes angle  with x-axis, i.e. = radian. While moving clockwise, here φ = +  Thus the x-projection of OP at time t will give us the equation of simple harmonic motion, given by

or x = -3 sin ωt (where x is in cm)
In figure
(b) T = 4s;A = 2m At t = 0, OP makes an angle π  with the positive direction of x-axis, i.e.  φ= π While moving anticlockwise, here φ=+π
Thus the x-projection of OP at time t will give us the equation of simple harmonic motion given by

Question 12.
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 
(b) x=cos 
(c) x=3sin 
(d) x=2 cos πt
Answer:

(d) Here, x = 2 cos πt
A = 2 cm, φ = 0 and ω= π rad s^-1
Therefore, at t = 0, the particle is at the point P on the right extreme position as shown in figure (iv).

Question 13.
Figure (i) 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 (ii) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in figure (ii) is stretched by the same force F.

(a) What is the maximum extension of the spring in the two cases?
(b) If the mass in figure (i) and the two masses in figure (ii) are released, what is the period of oscillation in each case?
Answer:
(a) Maximum extension of the spring :
(i) Suppose the maximum extension produced in the spring is y. Then,
F = ky   (in magnitude)
or y = F/k
(ii) In this case, each mass relative to the other behaves as if the other mass is fixed. In other words, force F on each mass acts as the force of reaction developed due to force F on the other mass. Therefore, in this case also, maximum extension is given by
y = F/k
(b) Period of oscillation:
In figure
(i), T = 2n T=2π√m/k
To calculate the period of oscillation, the spring in figure
(ii) can be considered as to be equivalent to the two springs, each of length 1/2 and joined at the point O, the center of the string as shown in the figure.

If k’ is force constant of each half, then k’ = 2k (Because, if a spring is cut to half of its length, its force constant becomes double. Therefore,
T = 2π√m/2k

Question 14.
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^-1, what is its maximum speed?
Answer:
Stroke of piston = 2 times the amplitude.
Let A = amplitude,
Stroke = 1 m (given)
∴ 1 =2A or A =  m
Now, v_max =  = 200 x  = 100 m/min
= ms^-1=  ms^-1

Question 15.
The acceleration due to gravity on the surface of moon is 1.7 m s^-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 m s^-2)
Answer:

Question 16.
Answer the following questions:
(a) Time period of a particle in S.H.M. depends on the force constant k and mass m of the particle :
T = 2π√ . A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
(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 2π√  Think of a qualitative argument to appreciate this result.
(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time under gravity?
(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?
Answer:
(a) For a simple pendulum, force constant or spring factor k is proportional to mass m, therefore, m cancels out in denominator as well as in numerator. That is why the time period of simple pendulum is independent of the mass of the bob.
(b) The effective restoring force acting on the bob of simple pendulum in displaced position is
F = -mgsinB. When θ is small, sin θ = θ . Then the expression for time period of simple pendulum is given by
T = 2π√ 
When θ is large sin θ < θ. If the restoring force mgsinθ is replaced by mgQ, this amounts to effective reduction in the value of ‘g’ for large angles and hence an increase in the value of time period T.
(c) Yes, because the working of the wrist watch depends on spring action and it has nothing to do with gravity.
(d) We know that gravity disappears under free fall, so frequency is zero.

Question 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:
When the car moves along a circular track, the bob of the simple pendulum will possess centripetal acceleration,
a_c=υ²/R  (radially inward along horizontal)
The acceleration due to gravity (g) acts vertically downwards.
Therefore, effective acceleration of the pendulum,

Question 18.
A cylindrical piece of cork of base area A and height h floats in a liquid of density p,. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period

where p is the density of cork. (Ignore damping due to viscosity of the liquid).
Answer:
Let X be the equilibrium position of a cylinder floating in a given liquid.
A = area of cross-section of the cylindrical piece of cork.
h = height of the cylindrical piece of cork,
ρ = density of the material of the cylindrical cork

ρ_c= density of the liquid in which it floats.
l = length of the cylindrical piece of cork dipping in the liquid upto point P in position X.
W = weight of the cylindrical cork.
W₁ = weight of the liquid displaced by the cork.
V = its volume m = mass of cork = Ahρ.
∴ V= Ah
W=mg = (Vρ)g = (Ahρ)g
W₁ = Area of cross-section of cork x length of cylinder dipping in liquid x density of liquid x g = Alρ₁g
According to the law of flotation,

Question 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 liquid column of mercury in the (U-tube executes simple harmonic motion.
Answer:
The suction pump creates the pressure difference, thus mercury rises in one limb of the U-Tube. When it is removed, a net force acts on the liquid column due to the difference in levels of mercury in the two limbs and hence the liquid column executes S.H.M. which can be explained as:

Consider the mercury contained in a vertical U-tube upto the level P and Q in its two limbs.
Let ρ = density of the mercury.
L = total length of the mercury column in both the limbs.
A = internal cross-sectional area of U-tube.
m = mass of mercury in U-tube = LAρ.
Let the mercury be depressed in left limb to P’ by small distance y, then it rises by the same amount in the right limb to position Q’
∴‍ Difference in levels in the two limbs
= P’Q’ = 2y
∴ Volume of mercury contained in the column of length
2y = A x 2y
m’ = A x 2y x ρ x g
If W = weight of liquid contained in the column of length 2y
Then W=m’g = A x 2y x ρ x g
This weight produces the restoring force (F) which tends to bring back the mercury to its equilibrium position.
∴ F = -2Ayρg = -(2Aρg)y
If a = acceleration produced in the liquid column, then

where h = height of mercury in each limb. Now from (i), it is clear that a ∝  y and -ve sign shows that it acts opposite to y, so the motion of mercury in U-tube is simple harmonic in nature having time period (T) given by

Question 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 as shown in figure. 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

Answer:
Consider an air chamber of volume V with a long neck of uniform area of cross section A, and a frictionless ball of mass m fitted smoothly in the neck at position C, the pressure of air below the ball inside the chamber is equal to the atmospheric pressure. Increase the pressure on the ball by a little amount P, so that the ball is depressed to position D, where CD = y.

There will be decrease in volume and hence increase in pressure of air inside the chamber. The decrease in volume of the air inside the chamber, ΔV = Δy

Clearly, F  ∝ y, Negative sign shows that the force is directed towards equilibrium position. If the applied increased pressure is removed from the ball, it will start executing linear SHM in the neck of chamber with C as mean position.
In SHM, the restoring force, F = -ky Comparing (i) and (ii), we have k = BA²/V, which is the spring factor.Now, inertia factor = mass of ball = m.
As,

Question 21.
You are riding an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of
(a) the spring constant k and
(b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.
Answer:
(a) Here mass supported by each wheel, M = 750 kg and x =15 cm = 0.15 m
If k is spring constant of the spring, then restoring force developed on being compressed through a distance x
F = -kx
If M is mass supported by each wheel, then
∴  k x = Mg

If b is damping constant for the spring and shock absorber system, then damped amplitude of oscillation is given by
A=A₀e-^bt/2M
where T is period of oscillation of the spring, A₀ the initial amplitude of the oscillations and  M, the mass supported by it. From the above relation we have
A=A₀e-^bt/2M
As the amplitude of oscillation decreases by 50% during one complete oscillation,


Question 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:
Consider a particle of mass m executing S.H.M. with period T. The displacement of the particle at an instant t, when time period is noted from the mean position is given by y = A sin ωt

Question 23.
A circular disc of mass 10 kg is suspended by a wire attached to its center. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant a is defined by the relation J = -αθ, where J is the restoring couple and 0 the angle of twist)
Answer:
Here, m = 10 kg; R = 15 cm ; T = 1.5 s Moment of inertia of the disc about wire
I =  MR² = –x 10 x (0.15)² = 0.1125 kg m²
When the wire is twisted by rotating the disc and then released, restoring torque will be set up. If a is angular acceleration produced, then restoring torque
τ = lα                                                         …(i)
The torsional spring constant k of the wire is defined by the relation
τ = -kθ                                                       …(ii)
where θ  is the angle of twist From the equations (i) and (ii) we have
lα= -kθ  or α= -k/l  θ
Since k/I is constant, it follows that angular acceleration is directly proportional to the angle of twist (angular displacement). Hence, the motion executed by the disc is simple harmonic in nature and the period of torsional oscillation is given by

Question 24.
A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of body when the displacement is
(a) 5 cm,
(b) 3 cm and
(c) 0 cm.
Answer:
Here, r = 5 cm = 0.05 m; T = 0.2 s;

Question 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 a distance x₀ and pushed towards the center with a velocity v₀ at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω ,x₀ and v₀.
Answer:
Let the displacement of the particle at any time t be represented by
x = Acos(ωt + φ_O)   …(i)
where A = amplitude, φ_O ⁼ initial phase If ν be the velocity of the particle at time t,

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