1st PUC Physics Question Bank Chapter 14 Oscillations

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Karnataka 1st PUC Physics Question Bank Chapter 14 Oscillations

Question 1.
What is oscillatory or vibrating motion?
Answer:
If a body moves back and forth repeatedly about a mean position at regular intervals of time, then the motion is called oscillatory or vibratory.

Question 2.
Give an example for oscillatory motion.
Answer:
A boat tossing up and down in a river, the piston in a steam engine going back and forth, motion of a loaded spring when the load is pulled down and released.

Question 3.
What is periodic motion?
Answer:
The motion which repeats itself again and again at regular intervals of time is called periodic motion.

Question 4.
Mention few example for periodic motion.
Answer:
Revolution of the earth around the sun, revolution of the moon around the earth, appearance of Halley Comet once in 76 years.

Question 5.
Define period.
Answer:
The smallest interval of time after which the motion is repeated is called period.
OR
The time taken by the particle to complete one oscillation is called period.

Question 6.
Define frequency.
Answer:
It is defined as the number of times the particle repeats its motion in one second.
OR
It is defined as the number of oscillation executed by the particle in one second.

Question 7.
Give the relation between period and frequency.
Answer:
Period = \(\frac{1}{\text { frequency }}\)
i.e., T = \(\frac{1}{v}\) or v = \(\frac{1}{T}\)

Question 8.
Mention the unit of frequency.
Ans: The unit of frequency is hertz (Hz) 1 Hz = 1s^-1.

Question 9.
Explain the term ‘linear displacement’.
Answer:
Consider a body executing periodic motion along a straight line about its mean position.
Example: Oscillations of a load spring.
The distance of the body from its mean position is called linear displacement (x)

Question 10.
Explain the term ‘angular displacement’.
Answer:
Consider a body executing periodic motion along the arc of a circle about its equilibrium position. Example: Oscillations of a simple pendulum. The angle made by the position vector with the vertical is called angular displacement (θ).

Question 11.
Mention the expression for displacement of a particle executing periodic motion.
Ans: In case of periodic motion, the displacement is a function of time. Therefore the displacement is given by
(i) f(t) = A cos ωt
(ii) f(t) = A sin ωt
(iii) f(t) = A cos ωt + B cos ωt

Question 12.
What is simple harmonic motion?
Answer:
It is an oscillatory motion which can be represented by a sine function or cosine function. The simplest form of periodic motion is known as simple harmonic motion.

Question 13.
Define simple harmonic motion.
Answer:
Simple harmonic motion is defined as the oscillatory motion in which the acceleration of the particle is directly proportional to the displacement and is directed towards the equilibrium position.

Question 14.
Mention the expression for displacement of a particle executing SHM and explain the terms.
Answer:
x(t) = A cos (ωt + Φ)
where x(t) = displacement x as a function of time.
A = amplitude
ω = angular frequency
(ωt + Φ) phase
Φ = phase constant

Question 15.
Write the equation of a particle executing SHM and represent the variation of displacement with time graphically.
Answer:
The equation of a particle SHM is x(t) = Acos (ωt + Φ)
As the cosine function varies from + 1 to – 1, displacement varies between + A and – A.

Question 16.
Show that SHM may be regarded as the projection of uniform circular motion along the diameter of a circle. Hence derive an expression for the displacement of a particle executing SRM.
Answer:
Consider a particle executing uniform circular motion in a circle of radius A with a constant angular velocity ω.
Let P be the position of the particle at any instant of the particle at any instant of time t. From P a projection M is drawn on the diameter XX’. As the particle describes circular motion, the projection M executes to and fro motion about the point O. Hence the projection of uniform circular motion executes simple harmonic motion.
Let θ be the angle made by the radius vector OP with the + X-axis.
from the triangle OPM,

cosθ = \(\frac{O M}{O P}\)
OM = OP cosθ
But OM = x, OP = A
x = A cosθ
But ω = \(\frac{\theta}{t}\) ⇒ θ = ωt
x = A cosωt
If Φ is the initial phase then x = A cos(ωt + Φ)

Question 17.
Define displacement of particle executing SHM.
Answer:
Displacement of a particle executing SHM at any instant is the distance of the particle from its mean position.

Question 18.
Define amplitude of SHM.
Answer:
Amplitude of a particle executing SHM is the maximum displacement of the particle from its mean position.

Question 19.
Define phase of particle executing SHM.
Answer:
Phase of a particle executing SHM is the state of vibration of the particle at that instant of time.

Question 20.
Define velocity of a particle executing SHM and obtain the expression for the same.
Answer:
The velocity of a particle executing SHM at any instant of time is the time rate of charge of displacement of the particle at that instant of time.
Partial velocity = \(\frac{\text { displacement }}{\text { time }}\)
v = \(\frac{d x}{d t}\)
But x(t) = Acos(ωt + Φ)

Question 21.
Where is the velocity of particle executing SHM (i) maximum (ii) minimum.
Answer:
Velocity of particle executing SIIM is v = \(-\omega \sqrt{A^{2}-x^{2}}\)
(i) Velocity of the particle is maximum when x = 0 i.e., the equilibrium position.
\(v_{\max }=-\omega \sqrt{A^{2}-0}=-\omega A\)
(ii) velocity of a particle is minimum when x = A i.e., at extreme position
\(v_{\min }=-\omega \sqrt{A^{2}-A^{2}}=0\)

Question 22.
Define acceleration of a particle SHM and obtain the expression for the same.
Answer:
The acceleration of a particle executing SHM at any instant of time is the time rate of change of velocity of particle at that instant of time.
Particle acceleration = \(\frac{\text { Change of velocity }}{\text { Time taken }}\)
a = \(\frac{d v}{d t}\)
v = -Aωsin(ωt + Φ)

Question 23.
Mention the expression for acceleration of a particle SHM and where the acceleration of the particle is
(i) maximum (ii) minimum.
Answer:
The acceleration of the particle SHM is a = -ω²x
(i) Acceleration of the particle is maximum when x = A i.e., when the particle is at extreme position,
a = -ω²A
(ii) Acceleration of the particle is minimum when x = 0 i.e., when the particle is at mean position.
a = 0

Question 24.
If the displacement of a particle executing SHM is x (t) = A cos ωt. Plot
(i) displacement – time graph (ii) velocity – time graph (iii) acceleration – time graph.
Answer:

Question 25.
Using force law (Hookes’ law) obtain an expression for angular frequency of a particle executing SHM. Ans: Consider a particle of mass m executing simple harmonic motion about the mean position. Let x be the displacement of the particle about its mean position.
The force acting on the particle is directly proportional to displacement and is in the opposite direction.
F ∝x
F = -kx ……..(1)
where k is a constant called force constant.
From Newtons II law of motion.
F = ma
But a = -ω²x
F = -mω²x ……(2)
From equation (1) and (2)
-kx = -mω²x
ω² = \(\frac{k}{m}\)
ω = \(\sqrt{\frac{k}{m}}\)

Question 26.
Obtain an expression for total energy of a particle executing simple harmonic motion.
Answer:
Consider a particle of mass m executing simple harmonic motion about its mean position. The vibrating particle has
(i) kinetic energy (K) due to motion of the particle.
(ii) potential energy U due to displacement of the particle from mean position.
∴total energy = kinetic energy + potential energy
E = K + U …….(1)
kinetic energy of the particle
K = \(\frac{1}{2} m v^{2}\)
But v = ω Asin(ωt + Φ)
K =\(\frac{1}{2}\)m[ωAsin(ωt + Φ)]²
K = \(\frac{1}{2}\) mω² Asin²(ωt + Φ)
But mω² = k, k is called force constant.
∴K = \(\frac{1}{2}\) kA²sin²(ωt + Φ) ……(2)
potential energy
U = \(\frac{1}{2}\) kx²
But x = A cos(ωt + Φ)
U = \(\frac{1}{2}\) kA²cos²(ωt + Φ) ……(3)
Substituting (2) and (3) in (1),

Question 27.
Where is the kinetic energy of a particle executing SHM (i) minimum (ii) maximum.
Answer:
(i) minimum at extreme position (ii) maximum at mean position.

Question 28.
Where is the potential energy of a particle executing SHM (i) minimum (ii) maximum.
Answer:
(i) minimum at mean position (ii) maximum at extreme position.

Question 29.
Give the graphical representation of the variation of potential energy, kinetic energy and the total energy as a function of position x for a particle executing SHM.
Answer:

Question 30.
Mention few characteristics (properties) of simple harmonic motion.
Answer:
The following are the important characteristics (properties) of SHM.

• The particle moves to and fro about the mean position.
• The displacement, velocity and acceleration vary sinusoidally with time but not in phase.
• The acceleration of the particle is proportional to the displacement but in opposite direction.
• The frequency and period of motion are independent of the amplitude of motion.

Question 31.
Obtain an expression for angular frequency and time period linear harmonic oscillator.
Answer:
Consider a block of mass fixed to a spring, which in turn fixed to a rigid wall as shown in the diagram.
If the block is pulled to one side and released, execute to and fro motion about mean position. Let the block oscillate between two extreme position.
Let x be the displacement of the block from its mean position at any instant of time t.
From Hookes law,
the restoring force ∝displacement from mean position.
F ∝x
F = -kx ……(1)
where k is a constant called spring constant.
From Newton’s II law F = ma
But a = -ω²x
∴F = -mω²x ……(2)
from (1) and (2)
-kx = -mω²x

Question 32.
Obtain an expression for the time period of an oscillating simple pendulum.
Answer:
Consider a simple pendulum i.e., a small bob of mass m tied to an inextensible string tied to a string of length L.
As the pendulum oscillates, let θ be the angle made by the string with the vertical.
There are two forces acting on the bob (i) tension T acting along the string (ii) force due to gravity mg acting vertically downwards.
The force mg can be resolved into two components as
(i) mg cosθ along the string
(ii) mg sinθ perpendicular to the string.
torque τ about the support due to the component mg sinθ is
torque = force × perpendicular distance
τ = -(mg sinθ) …….(1)
the negative sign is because the torque tends to reduce the angular displacement
Using Newton’s law for rotational motion.
τ = Iα ………(2)
where I is the moment of inertia and α is the angular acceleration
Comparing (1) and (2),
Iα = -mgL sinθ
for small angle sinθ ≅ 0
∴Iα = -mgLθ
α = \(\frac{-m g L}{I} \theta\) ……..(3)
equation (3) is similar to equation for acceleration for linear motion.
a = -ω²x ……..(4)
Comparing (3) and (4)
ω² = \(\frac{m g L}{I}\)

Question 33.
What is the length of a simple pendulum, which ticks seconds OR What is the length of the seconds pendulum?
Answer:
\(T=2 \pi \sqrt{\frac{L}{g}} \text { or } L=\frac{g T^{2}}{4 \pi^{2}}\)
The time period of a simple pendulum which ticks seconds is 2s.
\(L=\frac{9 \cdot 8 \times 4}{4 \pi^{2}}=1 \mathrm{m}\)

Question 34.
What is the time period of a seconds pendulum?
Answer:
Time period of seconds pendulum is 2s.

Question 35.
What are free oscillations? Give an example.
Answer:
The oscillations executed by a body when it is displaced from its equilibrium position and released are called free oscillations.
Example: Oscillations of a simple pendulum.

Question 36.
What is meant by natural frequency?
Answer:
The frequency of oscillations of a body executing free oscillation is called natural frequency.

Question 37.
What are damped oscillations? Give an example.
Answer:
If the amplitude of oscillations of a vibrating body decreases with time, the oscillations are called damped oscillations.
Example: Oscillations of a simple pendulum in the medium air.

Question 38.
Draw displacement time graph for damped oscillations.
Answer:

Question 39.
Mention the expression for angular frequency of a damped oscillator. Explain the terms.
Answer:
The angular frequency of a damped oscillator is
\(\omega=\sqrt{\frac{k}{m}-\frac{b^{2}}{4 m^{2}}}\)
where
ω = angular frequency
k = force constant
m = mass of the oscillator
b = a constant

Question 40.
What happens to amplitude of body executing damped oscillations?
Answer:
The amplitude decreases.

Question 41.
What happens to the mechanical energy of particle executing damped oscillations.
Answer:
The mechanical energy decreases.

Question 42.
What are damped oscillations?
Answer:
If the oscillations of a vibrating body die out with time due to dissipation of energy, the oscillations are called damped oscillations.

Question 43.
Mention an example for damped oscillations.
Answer:
Oscillations of simple pendulum in air.

Question 44.
What are forced oscillations?
Answer:
If a body is made to vibrate with the application of external periodic force, the body vibrates with the frequency of applied force, then the oscillations are called forced oscillations.

Question 45.
What is resonance?
Answer:
If the frequency of the applied force is gradually increased, the amplitude of the body and hence intensity of sound emitted also increases. When the frequency of the applied force is equal to the natural frequency of the body, the body vibrates with maximum amplitude. Consequently the intensity is also maximum. This phenomenon is called resonance.

Question 46.
What is the condition for resonance?
Answer:
The resonance taken place only when applied frequency is equal to natural frequency of the body.

Numerical Problems

Question 1.
On an average a human heart is found to beat 75 times in a minute. Calculate its frequency and period.
Answer:
The beat frequency of heart = 75/(1 min)
= 75 / (60 s)
= 1.25 s^-1
= 1.25 Hz
The time period T= 1/(1.25 s^-1) = 0.8 s

Question 2.
Which of the following functions of time represent (a) periodic and (b) non-periodic motion? Give the period for each case of periodic motion? [ω is any positive constant].
(i) sin ωt + cos ωt
(ii) sin ωt + cos 2 ωt + sin 4 ωt
(iii) e^-ax
(iv) log (ωt)
Answer:
(i) sin ωt + cos ωt is a periodic function, it can also be written as

The periodic time of the function is 2π/ω.
(ii) This is an example of a periodic motion. It can be noted that each term represents a periodic function with a different angular frequency. Since period is the least interval of time after which a function repeats its value, sin ωt has a period T₀ = 2π/ω; cos2ωt has a period π/ω = T₀/ 2; and sin 4 ωt has a period 2π/4ω = T₀/4. The period of the first term is a multiple of the periods of the last two terms. Therefore, the smallest interval of time after which the sum of the three terms repeats is To, and thus the sum is a periodic function with a period 2π/ω.
(iii) The function e^-ωt is not periodic, it decreases monotonically with increasing time and tends to zero as t → ∞ and thus, never repeats its value.
(iv) The function log(ωt) increases monotonically with time t. Therefore, it never repeats its value and is a non-periodic function. It may be noted that as t → ∞, log(ωt) diverges to ∞. Therefore, it cannot represent any kind of physical displacement.

Question 3.
Which of the following functions of time represent (a) simple harmonic motion and (b) periodic but not simple harmonic? Give the period for each case: (i) sin ωt – cos ωt (ii) sin ωt.
Answer:
(a) sin ωt – cos ωt

This function represents a simple harmonic motion having a period T = \(\frac{2 \pi}{\omega}\) and a phase angle \(\left(\frac{-\pi}{4}\right) \text { or }\left(\frac{7 \pi}{4}\right)\)
(b) sin²ωt
= \(\frac{1}{2}-\frac{1}{2} \cos 2 \omega t\)
The function is periodic having a period T = \(\frac{\pi}{\omega}\). It also represents a harmonic motion with the point of equilibrium occurring at \(\frac{1}{2}\) instead of zero.

Question 4.
A body oscillates with SHM according to the equation (in SI units). x = 5cos[2πt + \(\frac{\pi}{4}\)] At t = 1.5 s,
calculate the (a) displacement, (b) speed and (c) acceleration of the body.
Answer:
The angular frequency o of the body = 2πs’ arid its time period T = 1 s.
At t= 1.5 s.

= -5.0 × 0.707 m = – 3.535 m
(b) The speed of the body is given by
= -(5.0 m)(2π) sin\(\left[\left(3 \pi+\frac{\pi}{5}\right)\right]\)
= 10π × 0.707 ms^-1 = 22 ms^-1
(c) The acceleration of the body is given by
= -(2π)² × displacement
= -(2π)² × (-3.535)
= 140 ms^-2

Question 5.
A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 N m^-1. The block is pulled to a distance x = 10 cm from its equilibrium position at x = 0 on a frictionless surface from rest at t = 0. Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position.
Answer:
The block executes SHM, its angular frequency, as given by
ω = \(\sqrt{\frac{k}{m}}\)
\(\sqrt{\frac{50}{1}}=7.07 \mathrm{rad} \mathrm{s}^{-1}\)
Its displacement at any time t is then given by:
x(t) = 0.1 cos(7.07t)
Therefore, when the particle is 5 cm away from the mean position, we have
0.05 = 0.1 cos (7.07t)
Or
cos(7.07t) = 0.5
and hence sin(7.07t) = \(\frac{\sqrt{3}}{2}\) = 0.866
Then, the velocity of the block at x = 5 cm is
= 0.1 × 7.07 × 0.866 ms^-1 = 0.6 ms^-1
Hence the K.E. of the block,
= \(\frac{1}{2}\)mv²
= \(\frac{1}{2}\)[1 x (0.6123)²]=0.19J
The P.E. of the block,
= \(\frac{1}{2}\) kx²
= \(\frac{1}{2}\)(50 x 0.05 x 0.05) = 0.0625 J
The total energy of the block at x = 5 cm,
= K.E. + P.E. = 0.25 J
We also know that at maximum displacement, K.E. is zero and hence the total energy of the system is equal to the P.E. Therefore, the total energy of the system,
= \(\frac{1}{2}\)(50 × 0.1 × 0.1)= 0.25 J
which is same as the sum of the two energies at a displacement of 5 cm. This is in conformity with the principle of conservation of energy.

Question 6.
A 5 kg collar is attached to a spring of spring constant 500 Nm^-1. It slides without friction over a horizontal rod. The collar is displaced from the equilibrium position by 10.0 cm and released. Calculate (a) the period of oscillation, (b) the maximum speed and (c) maximum acceleration of the collar.
Answer:
(a) The period of oscillation as given by Eq. is,

(b) The velocity of the collar executing SHM is given by
v(t) = -Aωsin(ωt + Φ)
The maximum speed is given by,

(c) The acceleration of the collar at the displacement x (t) from the equilibrium is given by,
a(t) = -ω²x(t)
= \(-\frac{k}{m} x(t)\)
Therefore the maximum acceleration is,
a_max = ω²A
= \(\frac{500}{5} \times 0 \cdot 1 \mathrm{m}=10 \mathrm{ms}^{-2}\) and it occurs at thee extremities.

Question 7.
What is the length of a simple pendulum, which ticks seconds?
Answer:
From Eq. the time period of a simple pendulum is given by
T= \(2 \pi \sqrt{\frac{L}{g}}\)
From this relation one gets,
L = \(\frac{g T^{2}}{4 \pi^{2}}\)
The time period of a simple pendulum, which ticks seconds, is 2 s. Therefore for g =98 ms^-2 and T = 2s, L is
= \(\frac{9 \cdot 8 \times 4}{4 \pi^{2}}\) = 1m

Question 8.
For the damped oscillator, the mass m of the block is 200 g, k = 90 Nm and the damping constant b is 40 g s^-1. Calculate (a) the period of oscillation, (b) time taken for its amplitude of vibrations to drop to half of its initial value of (c) the time taken for its mechanical energy to drop to half its initial value.
Answer:
(a) We see that k × m = 90 × 0.2 = 18 kg Nm^-1 = kg²s^-2, t, therefore, \(\sqrt{k m}\) = 4.243 kg s^-1, and b = 0.04 kg s^-1. Therefore b is much less than \(\sqrt{k m}\). Hence the time period T is given by

(b) The time, T_1/2, for the amplitude to drop to half of its initial value is given by,

(c) For calculating the time, t_1/2, for its mechanical energy to drop to half its initial value we make use of the following equation,

This is just half of the decay period for amplitude.

Question 9.
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:
(b) and (c)

Question 10.
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.
(h) motion of an oscillating mercury column in a U-tube.
(c) motion of a bail bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.
(d) general vibrations of a poly-atomic molecule about its equilibrium position.
Answer:
(a) Periodic but not SHM (Simple Harmonic Motion)
(b) SHM
(c) SHM
(d) Periodic but not SHM

Question 11.
Figure below 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:
(b) and (d) are in periodic motion with period of 2 sec.

Question 12.
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is
(a) at the end A,
(b) at the end B,
(c) at the mid-point of AB going towards A,
(d) at 2 cm away from B going towards A,
(e) at 3 cm away from A going towards B, and
(f) at 4 cm away from B going towards A.
Ans:

Question 13.
Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?
(a) a = 0.7 x
(b) a = – 200 x²
(c) a = – 10 x
(d) a = 100 x².
Answer:
a = – 10x (∵acceleration is proportional and opposite to displacement)

Question 14.
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 1 cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is πs^-1. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.
Answer:
Given: x(t) = A cos(ωt + Φ) …….(1)
at t = 0, x(0) = 1 cm, ω = πs^-1
Substituting in (1) we get
1 cm = Acos(0 x π + Φ)
AcosΦ = 1 ……(2)
Differentiate equation (1) w.r.t. t
\(\frac{d x}{d t}-A \omega \sin (\omega t+\phi)\) ………(3)
At t = 0, v = ω cm/s, ω = πs^-1
Substituting in 3 we get
ω = -AωsinΦ
1 = —A sinΦ ……(4)
Divide equation (4) by (2) we get
1 = \(\frac{-\sin \phi}{\cos \phi}\)
⇒tan Φ = -1
Φ = \(\frac{-\pi}{4}\)
⇒Φ = initial phase
Substitute Φ value in equation (2) we get
1 = \(A \cos \left(\frac{-\pi}{4}\right)\)
⇒ A = \(\sqrt{2}\) cm
If x(t) = Bsin(ωt + α) ………(5)
at t = 0, x = 1 cm, ω = πs^-1
substituting in (5) we get
1 = Bsin (α) ………(6)
Differentiate (5) w.r.t. ‘t’
⇒ v = \(\frac{d x}{d t}\) = Bω cos(ωt + α)
at t = 0, v = ω = π
⇒ 1 = Bcos(α) ……..(7)
Divide (7) by (6)
we get tan Φ = 1
⇒ Φ = \(\frac{\pi}{4}\) is the initial phase
Substitute Φ in equation (6) we get
1 = Bsin(\(\frac{\pi}{4}\)) ⇒ B = \(\sqrt{2}\) cm

Question 15.
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:
Maximum length Ymax = 0.2 m
Maximum weight Mmsa = 50 kg
We know that, F = Mg – ky, at M = Mmax, Y = Y

Weight of the body = 22.36 x 9.8 = 219.1 N

Question 16.
A spring having a spring constant 1200 N m^-1 is mounted on a horizontal table as shown in Fig. 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, (ii) maximum acceleration of the mass and (iii) the maximum speed of the mass.

Answer:
Given: K = 1200 Nm-1, m = 3 kg, a = 2 cm
(i) Frequency of oscillation
we know that T = \(2 \pi \sqrt{\frac{m}{K}}=2 \pi \sqrt{\frac{3}{1200}}\)
T = 0.3144 s
f = \(\frac{1}{T}\) = 3.18/s

(ii) Maximum acceleration of mass (α max) is obtained when y = a

(iii) Maximum speed of the mass

Question 17.
In exercise 17, 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 SHM differ from each other, in frequency, in amplitude or the initial phase?
Answer:
Given a = 2 cm, k = 1200 Nm^-1, m = 3 kg
\(\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{1200}{3}}=20 \mathrm{s}^{-1}\)
Let equation of SHM, x = A sin ωt
(a) time is measured from mean position.
x = A sin ωt
x – 2 sin 20t
(b) at maximum structured position phase angle is \(\frac{\pi}{2}\)
x = asin (ωt + \(\frac{\pi}{2}\))
x = 2cos(20t)
(c) At maximum compressed position phase angle is \(\frac{3 \pi}{2}\)
x = asin (ωt + \(\frac{3 \pi}{2}\))
x = -acos ωt

Question 18.
Figure below 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:

(a) Let A be any point on the circle show in fig (a). Draw AM⊥L to x – axis. Point M refers to x – projection of the radius vector. Now,

(∵Alternate angles)

(b) let A be any point on the circles of fig. (b) From A draw AM ⊥ to x-axis.

Question 19.
Plot the cor responding 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 anti-clockwise In every case: (x is in cm and t is in s).

Answer:

Question 20.
Figure (a) shows a spring of force constant k damped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. (b) is stretched by the same force F.

(a) What is the maximum extension of the spring in the two cases?
(b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?
Answer:
(a) Consider figure (a)
Let y be the extension produced in the spring
F = ky
y = \(\frac{F}{k}\)
Consider figure (b)
Each mass acts as if it is fixed w.r.t. the other
F = ky
y = \(\frac{F}{k}\)
(b) Consider figure (a)
F = -ky


Let us assume (1) as centre of system and 2 springs each of length \(\frac{1}{2}\) attached to two masses. So k’ is the spring factor of each spring.
k’ = 2k
⇒T = \(\frac{2 \pi}{\omega}\)
T = \(2 \pi \sqrt{\frac{m}{2 k}}\)
is the period of oscillation in the case of (b)

Question 21.
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad / min, what is its maximum speed?
Answer:
Stroke = 1 m

Question 22.
The acceleration due to gravity on the surface of moon as 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:
g(1) =1.7 ms^-2 , g_e =9.8ms^-2, T_e = 3.5 s

Question 23.
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 = \(2 \pi \sqrt{\frac{m}{k}}\). A simple pendulum executes SHM approximately. Why then is the 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 \(2 \pi \sqrt{\frac{I}{g}}\) Think of a qualitative argument to appreciate this result.
(c) A man with a wristwatch on his hand fails from the top of a tower. Does the watch give correct time during the free fall?
(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely failing under gravity?
Answer:
(a) In case of a simple pendulum k is directly proportional to m. Hence ratio of m / k is a constant. Hence time period doesn’t depend on mass, (b) Restoring force that brings bob of pendulum back to its mean position
F = -mg sinθ
For small

If approximate for sinθ = θ is not taken into account then time period T > \(2 \pi \sqrt{\frac{1}{g}}\)
This happens when θ is not small.
(c) Wrist watch work on the principle of spring action. Hence acceleration due to gravity plays no role in the functioning of wrist watch hence it gives correct time during free fall.
(d) During free fall acceleration due to gravity is zero hence the pendulum will not vibrate (i.e., frequency of oscillation is zero).

Question 24.
A simple pendulum of length 1 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:
Bob of the pendulum is under the action of two accelerations, acceleration due to gravity ‘g’ and centripetal

Question 25.
A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρ₁. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period T = \(2 \pi \sqrt{\frac{h \rho}{\rho_{1} g}}\) where p is the density of cork. (Ignore damping due to viscosity of the liquid).
Answer:
Initially in equilibrium, weight of cork = weight of water displaced \(h \rho=h^{\prime} \rho_{e}\)
\(h=\frac{h^{\prime} \rho_{e}}{\rho}\)
When cork is pushed in, then the restoring force acting on it is:
f = – weight of the portion dipped after pushing
= -(Ay)ρ_eg

Now,

Question 26.
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:
Restoring force acting on the liquid when suction pump is removed is,
f = -mg
f = -(A × 2y)ρ × g
where A = cross – section area of U tube ρ = density of liquid
f =-2ρgy
acceleration produced in liquid column,

But L = 2h

which is of form a = -ω²x
ω = \(\sqrt{\frac{g}{h}}\)
Hence it is a SHM.

Question 27.
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. 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:
Volume = V, mass of ball = m, cross-sectional area = A
Initial pressure on either side of the ball = atmospheric pressure = P
Let the charge in volume of air when ball is pressed be ΔV
ΔV = Ay(y = displacement)

Now, Bulk modulus of elasticity



Hence it is in SHM.

Question 28.
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 particle executing SHM starts from mean position.
Displacement can be given by
x = A sinωt
Velocity v = Aω cosωt
Kinetic energy over one complete cycle

Question 29.
A circular disc of mass 10 kg is suspended by wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation J = – αθ, where J is the restoring couple and θ the angle of twist).
Answer:
We know that,

Question 30.
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 the body when the displacement is (a) 5 cm, (b) 3 cm, (c) 0 cm.
Answer:
Given, r = 5cm, T = 0.2 .
\(\omega=\frac{2 \pi}{T}=\frac{2 \pi}{0 \cdot 2}\) = 10π rad/s
Acceleration A = -ω²y
Velocity V = \(\omega \sqrt{r^{2}-y^{2}}\)
(a) y = 5 cm
A = -(10π)² × 0.05
⇒ A = 0.493 m/s²
V = \(10 \pi \sqrt{(0 \cdot 05)^{2}-(0 \cdot 05)^{2}}=0\)
(b) y = 3 cm
A = (10π)² × 0.03
⇒ A = -0.296 m / s²
V = \(10 \pi \sqrt{(0 \cdot 05)^{2}-(0 \cdot 03)^{2}}\)
V = 0.4 πm/s
(c) y = 0 cm, A = 0
V = \(10 \pi \sqrt{(0 \cdot 05)^{2}-0}\)
= 0.5 πm/s

Question 31.
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 centre with a velocity v₀ at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ra, x₀ and v₀.
(Hint: Start with the equation x = a cos (ωt + θ) and note that the initial velocity is negative.]
Answer:
Let x = a cos(ωt + θ)
∴v = \(\frac{d x}{d t}\) =-aωsin((ωt + θ))
When t = 0, x = x₀ and \(\frac{d x}{d t}\) = -v₀
⇒ x₀ = a cosθ …….(1) and
-v₀ = -aω sinθ
⇒a sinθ = \(\frac{v_{0}}{\omega}\) ……..(2)
Squaring and adding (1) and (2, a^2x)
\(=x_{0}^{2}+\frac{v_{0}^{2}}{\omega^{2}} \Rightarrow a=\sqrt{x_{0}^{2}+\frac{v_{0}^{2}}{\omega^{2}}}\)