TS Inter 1st Year Physics Study Material Chapter 8 Oscillations

Telangana TSBIE TS Inter 1st Year Physics Study Material 8th Lesson Oscillations Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 8th Lesson Oscillations

Very Short Answer Type Questions

Question 1.
Give two examples of periodic motion which are not oscillatory.
Answer:

1. Motion of seconds hand of a watch.
2. Motion of fan blades which are rotating with constant angular velocity ‘w’.

For these two cases, they have constant centrifugal acceleration which does not change with rotation so it is not considered

Question 2.
The displacement in S.H.M. is given by y = a sin (20t + 4). What is the displacement when it is increased by2π/ω?
Answer:
Displacement :
Displacement remains constant ; \(\frac{2 \pi}{\omega}\) = time period T. After a time (0
period T, there is no change in equation of S.H.M.
i.e. Y = A sin (20t + 4) = Y = A sin (201 + 4 + T)
∴ There is no change in change in displacement.

Question 3.
A girl is swinging seated in a swing. What is the effect on the frequency of oscillation if she stands?
Answer:
The frequency of oscillation (n) will increase because in the standing position, the location of centre of mass of the girl shift upwards. Due to it, the effective length of the swing decreases. As n ∝ \(\frac{1}{\sqrt{l}}\), therefore, n increases.

Question 4.
The bob of a simple pendulum is a hollow sphere filled with water. How will the period of oscillation change, if the water begins to drain out of the hollow sphere?
Answer:
When water begins to drain out of the sphere, the centre of mass of the system will first move down and then will come up to the initial position. Due to this the equivalent length of the pendulum and hence time period first increases, reaches a maximum value and then decreases till it becomes equal to its initial value.

Question 5.
The bob of a simple pendulum is made of wood. What will be the effect on the time period if the wooden bob is replaced by an identical bob of aluminum?
Answer:
The time period of a simple pendulum does not change, if the wooden bob is replaced by an identical bob of aluminium because the time period of a simple pendulum is independent of the material of the bob.

Question 6.
Will a pendulum clock gain or lose time when taken to the top of a mountain?
Answer:
At higher altitudes i.e., on mountains the acceleration due to gravity is less as compared on the surface of earth. Since time period is inversely proportional to the square root of the acceleration due to gravity, the time period increases. The pendulum clock loses time on the top of a mountain.

Question 7.
What is the length of a simple pendulum which ticks seconds? (g = 9.8 ms^-2) [AP Mar. ’18: TS Mar. ’15]
Answer:
In simple pendulum T = 2π\(\sqrt{\frac{l}{g}}\) or l = \(\sqrt{\frac{gt^2}{4\pi^2}}\)
For seconds pendulum T = 2s ⇒ t² = 4
∴ = \(\frac{9.8\times4}{4\pi^2}\) = 1 m (∴ π² nearly 9.8)

Question 8.
What happens to the time period of a simple pendulum if its length is increased upto four times?
Answer:
In simple pendulum Time period T ∝ √l

From the above equation time period is doubled.

Question 9.
A pendulum clock gives correct time at the equator. Will it gain or lose time if it is taken to the poles? If so, why?
Answer:
When a pendulum clock showing correct time at equator is taken to poles then it will gain time.

Acceleration due to gravity at poles is high. Time period of pendulum T = 2π\(\sqrt{\frac{l}{g}}\).

When g increases T decreases. So number of oscillations made in the given time increases hence clock gains time.

Question 10.
What fraction of the total energy is K.E when the displacement is one half of a amplitude of a particle executing S.H.M?
Answer:
Kinetic energy is equal to three fourth (i.e.,\(\frac{3}{4}\)) of the total energy, when the displacement is one-half of its amplitude.

Question 11.
What happens to the energy of a simple harmonic oscillator if its amplitude is doubled?
Answer:
Energy of a simple harmonic oscillator,

From the above equation, energy increases by four times.

Question 12.
Can a simple pendulum be used in an artificial satellite? Give the reason
Answer:
No, this is because inside the satellite, there is no gravity, i.e., g = 0. As T = 2π\(\sqrt{\frac{l}{g}}\) where T = ∞ for g = 0. Thus, the simple pendulum will not oscillate.

Short Answer Questions

Question 1.
Define simple harmonic motion? Give two examples.
Answer:
Simple Harmonic Motion :
A body is said to be in S.H.M, if its acceleration is directly proportional to its displacement, acts opposite in direction towards a fixed point.

Examples:

1. Projection of uniform circular motion on a diameter.
2. Oscillations of simple pendulum with small amplitude.
3. Oscillations of a loaded spring.
4. Vibrations of a liquid column in U – tube.

Question 2.
Present graphically the variations of displacement, velocity and acceleration with time for a particle in S.H.M.
Answer:
The variations of displacement, velocity and acceleration with time for a particle in S.H.M can be represented graphically as shown in the figure.

From the graph

1. All quantities vary sinusoidally with time.
2. only their maxima differ and the different plots differ in phase.
3. Displacement x varies between – A to A; v(t) varies from – ωA to ωA and a (t) varies from – ω²A to ω²A.
4. With respect to displacement plot, velocity plot has a phase difference of \(\frac{\pi}{2}\) and acceleration plot has a phase difference of π.

Question 3.
What is phase? Discuss the phase relations between displacement, velocity and acceleration in simple harmonic motion.
Answer:
Phase (θ) :
Phase is defined as its state or condition as regards its position and direction of motion at that instant.
In S.H.M phase angle, θ = ωt = 2π(\(\frac{t}{T}\))

a) Phase between velocity and displacement :
In S.H.M, displacement,
y = A sin (ωt – Φ)
Velocity, V = Aω cos (ωt – Φ)
So phase difference between displacement and velocity is 90°.

b) Phase between displacement and acceleration :
In S.H.M, acceleration ‘a’ = – ω²y
or y = A sin ωt and a = – ω² A sin ωt
– ve sign indicates that acceleration and displacement are opposite.

So phase difference between displacement and acceleration is 180°.

Question 4.
Obtain an equation for the frequency of oscillation of spring of force constant k to which a mass m is attached.
Answer:
Let a spring of negligible mass is suspended from a fixed point and mass m is attached as shown. It is pulled down by a small distance ‘x’ and allowed free it will execute simple harmonic oscillations.

Displacement from mean position = x.
The restoring forces developed are opposite to displacement and proportional to ‘x’. ∴ F ∝ – x or F = – kx where k is constant of spring, (-ve sign for opposite direction)

Question 5.
Derive expressions for the kinetic energy and potential energy of a simple harmonic oscillator.
Answer:
Expression for K.E of a simple harmonic oscillator :
The displacement of the body in S.H.M., X = A sin ωt
where A = amplitude, ωt = Angular displacement.

Velocity at any instant, v = \(\frac{dx}{dt}\) = Aω cos ωt
∴ K.E = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\)mA²ω² cos² ωt

At mean position velocity is maximum and displacment x = 0
∴ K.E_max = \(\frac{1}{2}\)mA²ω²

Expression for P.E of a simple harmonic oscillator:
Let a body of mass m’ is in S.H.M with an amplitude A.

Let O is the mean position.
Equation of a body in S.H.M is given by, x = A sin ωt
For a body in S.H.M acceleration, a = – ω²Y
Force, F = ma = – mω²x
∴ Restoring force, F = mω²x

Potential energy of the body at any point say ‘x’ :
Let the body is displaced through a small distance dx
Work done, dW = F . dx = P.E.
This work done.
∴ P.E = mω²x. dx(where x is its displacement)
Total work done, W = ∫dW = \(\int_0^x m \omega^2 x\).dx
⇒ Work done, W = \(\frac{m\omega^2x^2}{2}\).
This work is stored as potential energy.
∴ P.E at any point = \(\frac{1}{2}\)mω²x²

Question 6.
How does the energy of a simple pendulum vary as it moves from one extreme position to the other during its oscillations?
Answer:
The total energy of a simple pendulum is,
E = \(\frac{1}{2}\)mA² (or) E = \(\frac{1}{2}\frac{mg}{l}\)A²

The above equation, shows that the total energy of a simple pendulum remains constant irrespective of the position at any time during the oscillation i.e., the law of conservation of energy is valid in the case of a simple pendulum. At the extreme positions P and Q the energy is completely in the form of potential energy and at the mean position 0 it is totally converted as kinetic energy.

At any other point the sum of the potential and kinetic energies is equal to the maximum kinetic energy at the mean position or maximum potential energy at the extreme position. As the bob of the pendulum moves from P to O, the potential energy decreases but appears in the same magnitude as kinetic energy. Similarly as the bob of the pendulum moves from 0 to P or Q, the kinetic energy decreases to the extent it is converted into potential energy, as shown in figure.

Question 7.
Derive the expressions for displacement, velocity and acceleration of a particle executes S.H.M.
Answer:
Displacement of a body in S.H.M.
X = A cos (ωt + Φ).

i) Displacement (x) :
At t = 0 displacement x = A i.e., at extreme position when ωt + Φ = 90° displacement x = 0 at mean position at any point x = A cos (ωt + Φ).

ii) Velocity (V): Velocity of a body in S.H.M.

When (ωt + Φ) = 0 then velocity v = 0. For points where (ωt + Φ) = 90°
Velocity V = – Aω i.e., velocity is maximum,

iii) Acceleration (a): Acceleration of a body in S.H.M. is a = \(\frac{dv}{dx}\)
= \(\frac{d}{dt}\)(-Aω sin(ωt + Φ) = -Aω²cos(ωt + Φ) = -ω²x)
a_max = -ω²A

Long Answer Questions

Question 1.
Define simple harmonic motion. Show that the motion of projection of a particle performing uniform circular motion, on any diameter is simple harmonic. [TS May 18, Mar. 16, June 15; AP Mar. ;19, 18, AP May 16, 14]
Answer:
Simple harmonic motion :
A body is said to be in S.H.M, if its acceleration is directly proportional to its displacement, acts opposite in direction towards a fixed point.

Relation between uniform circular motion and S.H.M.:
Let a particle ‘P’ is rotating in a circular path of radius ‘ω’ with a uniform angular velocity ‘P’. After time ‘t’ it goes to a new position ‘P’. Draw normals from ‘P’ on to the X – axis and on to the Y – axis. Let ON and OM are the projections on X and Y axis respectively.

As the particle is in motion it will subtend an angle θ = ωt at the centre.

From triangle OPN
ON = OP cos θ
But OP = r and θ = ωt
∴ Displacement of particle P on X – axis at any time t is
X = r cos ωt ………… (1)
From triangle OPM
OM = Y = OP sin θ
But OP = r and θ = ωt
∴ Displacement of particle P on Y- axis is
Y = r sin ωt ………… (2)

As the particle rotates in a circular path the foot of the perpendiculars OM and ON will oscillate with in the limits X to X¹ and Y to Y¹.

At any point the displacement of particle P is given by OP² = OM² = ON²
Since OM = X = r cos ωt and ON = Y = r sin ωt.

So a uniform circular motion can be treated as a combination of two mutually perpendicular simple harmonic motions.

Question 2.
Show that the motion of a simple pendulum is simple harmonic and hence derive an equation for its time period. What is a seconds pendulum? [TS Mar. 18, 17, 15, May 17, 16; AP Mar. 17, 16. 15, 14, 13; AP May 18. 17. 13; June 15]
Answer:
Simple pendulum :
Massive metallic bob is suspended from a rigid support with the help of inextensable thread. This arrangement is known as simple pendulum.

So length of simple pendulum is ‘l’. Let the pendulum is pulled to a side by a small angle ‘θ’ and released it oscillate about the mean position.

Let the bob is at one extreme position B. The weight (W = mg) of body acts vertically downwards.

By resolving the weight into two perpendicular components :

1. One component mg sin θ is responsible for the to and fro motion of pendulum.
2. Other component mg cos θ will balance the tension in the string.

Force useful for motion F = mg sin θ = ma (From Newton’s 2nd Law)
From the above equations
∴ a = g sin θ

Since acceleration is proportional to displacement and acceleration is always directed towards a fixed point the motion of simple pendulum is “simple harmonic”.

Time period of simple pendulum :

Seconds pendulum :
A pendulum whose time period is 2 seconds is called “seconds pendulum.”

Question 3.
Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion is constant at any point on its path.
Answer:
Expression for K.E of a simple harmonic oscillator :
The displacement of the body in S.H.M, X = A sin ωt
where A = amplitude and ωt = Angular displacement.

Velocity at any instant, v = \(\frac{dx}{dt}\) = Aω cos ωt

At mean position velocity is maximum and displacement x = 0
∴ K.E_max = \(\frac{1}{2}\)mA²ω²

Expression for P.E of a simple harmonic oscillator :
Let a body of mass’m’ is in S.H.M with an amplitude A.

Let O is the mean position.
Equation of a body in S.H.M is given by, x = A sin ωt
For a body in S.H.M acceleration, a = – ω²Y
Force, F = ma = – mω²x
∴ Restoring force, F = mω²x

Potential energy of the body at any point say ‘x’:
Let the body is displaced through
a small distance dx
⇒ Work done, dW = F . dx
This work done = RE. in the body
∴ P.E = mω²x. dx(where x is its displacement)
Total work done, W = ∫dW = \(\int_0^x m \omega^2 x\).dx
work done, W = \(\frac{m\omega^2 x^2}{2}\)
This work is stored as potential energy.
∴ P.E at any point = \(\frac{1}{2}\)ω²x²
For conservative force total Mechanical Energy at any point = E= P.E + K.E
∴ Total energy,
E = \(\frac{1}{2}\)mω²(A² – x²) + \(\frac{1}{2}\)mω²x²
E = \(\frac{1}{2}\)mω²{A² – x² + x²} = \(\frac{1}{2}\)mω²A²

So for a body in S.H.M total energy at any point of its motion is constant and equals to \(\frac{1}{2}\)mω²A²

Problems

Question 1.
The bob of a pendulum is made of a hollow brass sphere. What happens to the time period of the pendulum, if the bob is filled with water completely? Why?
Solution:
If the hollow brass sphere is completely filled with water, then time period of simple pendulum does not change. This is because time period of a pendulum is independent of mass of the bob.

Question 2.
Two identical springs of force constant “k” are joined one at the end of the other On series). Find the effective force constant of the combination.
Solution:
When two springs of constant k each are joined together with end to end in series then effective spring constant k = \(\frac{k_1k_2}{k_1+k_2}\) in this case k_eq = \(\frac{k.k}{k+k}=\frac{k}{2}\)

In series combination, force constant of springs decreases.

Question 3.
What are the physical quantities having maximum value at the mean position in SHM?
Solution:
In S.H.M at mean position velocity and kinetic energy will have maximum values.

Question 4.
A particle executes SHM such that, the maximum velocity during the oscillation is numerically equal to half the maximum acceleration. What is the time period? [TS June ’15]
Solution:
Given maximum velocity, V_max = \(\frac{1}{2}\) maximum acceleration (a_max)
But V_max = Aw and a_max = ω² A
∴ Aω = \(\frac{1}{2}\) . Aω² ⇒ ω = 2
Time period of the body, T = \(\frac{2 \pi}{\omega}=\frac{2 \pi}{2}\)

Question 5.
A mass of 2 kg attached to a spring of force constant 260 Nm^-1 makes 100 oscillations. What is the time taken?
Solution:
Mass attached, m = 2 kg ; Force constant, k = 260 N/m
∴ Time period of loaded spring, T = 2π\(\sqrt{\frac{m}{k}}\)
= 2π\(\sqrt{\frac{2}{260}}\) = 0.5509 sec
∴ Time for 100 oscillations = 100 × 0.551
= 55.1 sec

Question 6.
A simple pendulum in a stationary lift has time period T. What would be the effect on the time period when the lift (i) moves up with uniform velocity (ii) moves down with uniform velocity (iii) moves up with uniform acceleration ‘a’ (iv) moves down with uniform acceleration ‘a’ (v) begins to fall freely under gravity?
Solution:
i) When the lift moves up with uniform velocity i.e., a = 0, there would be no change in the time period of a simple pendulum.

ii) When the lift moves down with uniform velocity i.e., a = 0, there would be no change in the time period of a simple pendulum.

iii) When lift is moving up with acceleration ‘a’ then relative acceleration = g + a
∴ Time period, T = 2 π\(\sqrt{\frac{l}{g+a}}\) so when lift is moving up with uniform acceleration time period of pendulum in it decreases.

iv) When lift is moving down with acceleration ‘a’ time period, T = 2π\(\sqrt{\frac{l}{g-a}}\)
(g – a = relative acceleration of pendulum)
So time period of pendulum in the lift decreases.

v) If the lift falls freely, a = g then the time period of a simple pendulum becomes infinite.

Question 7.
A particle executing SHM has amplitude of 4cm, and its acceleration at a distance of 1cm from the mean position is 3cms^-2. What will its velocity be when it is at a distance of 2cm from its mean position?
Solution:
Amplitude, A = 4cm = 4 × 10^-2m
Acceleration, a = 3cm/s² = 3 × 10^-2 m/s²;
Displacement, y = 1cm = 10^-2 m
∴ Angular velocity, ω = \(\sqrt{\frac{a}{y}}=\sqrt{\frac{3}{1}}=\sqrt{3}\)

To find velocity at a displacement of 2cm

Question 8.
A simple harmonic oscillator has a time period of 2s. What will be the change in the phase after 0.25 s after leaving the mean position?
Solution:
Time period, T = 2 sec; time, t = 0.25 sec
Phase difference after t sec = Φ = \(\frac{t}{T}\) × 2π
= \(\frac{0.25}{2}\) × 2π = \(\frac{2 \pi}{4}\) = 90°
For a phase of \(\frac{2 \pi}{4}\) starting from mean position the body will be at extreme position. (Phase difference between mean position and extreme position is \(\frac{2 \pi}{4}\) Rad or 90°)

Question 9.
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.
Solution:
Given that, A = 5 cm = 5 × 10^-2m and T = 0.2 s
Angular velocity, ω = \(\frac{2 \pi}{T}=\frac{2 \pi}{0.2}\) = 10π rad s^-1

a) Displacement, y = 5 cm = 5 × 10^-2 m
i) Acceleration of the body, a = – ω²y
= -(10π)² × 5 × 10^-2 = -5π²ms^-2
ii) Velocity of the body,

b) Displacement, y = 3 cm = 3 × 10^-2 m
i) Acceleration of the body, a = – ω²y
= -(10π)² × 3 × 10^-2 = -3π²ms^-2
ii) Velocity of the body, v = ω\(\sqrt{A^2 – y^2}\)

= 10π × 4 × 10^-2 = 0.4π ms^-1

c) Displacement, y = 0 cm
i) Acceleration of the body, a = – ω²y = 0
ii) Velocity of the body, v = ω\(\sqrt{A^2 – y^2}\)
= 10π^(5xl0’2)2-(0)2\(\sqrt{(5\times10^{-2})^2-(0)^2}\)
= 10π × 5 × 10^-2 =0.5π ms^-1

Question 10.
The mass and radius of a planet are double that of the earth. If the time period of a simple pendulum on the earth is T, find the time period on the planet.
Solution:
Mass of planet, M_P = 2 M_e ;
Radius of planet, R_P = 2R_e
Time period of pendulum on earth = T ;
Time period on planet = T’

Question 11.
Calculate the change in the length of a simple pendulum of length lm, when its period of oscillation changes from 2 s to 1.5 s. [TS Mar. ’18]
Solution:
For seconds pendulum T₁ = 2 sec ;
Length l₁ = 1 m.
New time period T₂ = 1.5 sec; Length l₂ = ?

Question 12.
A freely falling body takes 2 seconds to reach the ground on a plane, when it is dropped from a height of 8m. If the period of a simple pendulum is seconds on the planet. Calculate the length of the pendulum.
Solution:
Height, h = 8m;
Time taken to reach the ground, t = 2 sec
But for a body dropped, t = \(\sqrt{\frac{2h}{g}}\)
⇒ 2 = \(\sqrt{\frac{16}{g}}\) ⇒ g = \(\frac{16}{4}\) = 4m/s² on that planet
Time period of pendulum, T = 2π\(\sqrt{\frac{l}{g}}\) = π
∴ 2\(\sqrt{\frac{l}{g}}\) = 1 or \(\frac{l}{g}=\frac{1}{4}\) ⇒ l = \(\frac{g}{4}\)
Length of pendulum = \(\frac{4}{4}\) = 1m = 100cm on that planet

Question 13.
Show that the motion of a simple pendulum is simple harmonic and hence derive an equation for its time period.
Find the length of a simple pendulum which ticks seconds, (g = 9.8 ms^-2) [AP Mar. ’18. ’16, ’15, May ’17, June ’15; TS Mar.’17, 15, May 17]
Solution:
Simple pendulum :
In a laboratory a heavy metallic bob is suspended from a rigid support with the help of a spunless thread. This arrangement is known as “simple pendulum”.

Let the length of simple pendulum is ‘l’ and the point of suspension is ‘S’. Let the pendulum is drawn to a side by a small angle ‘θ’ and allowed free to oscillate in the vertical plane. Then it will oscillate between the extreme positions A and B with a displacement say ‘x’ at any given time.

Let the bob is at one extreme position say B. The force vertically acting downwards is Weight W = mg.

By resolving the weight into two per-pendicular components:

1. The component mg sin θ is responsible for the to and fro motion of the bob.
2. The component mg cos θ will balance the tension in the string.

Force useful for motion F = mg sin θ
= ma (From Newton’s 2nd Law)
∴ a = g sin θ

Since acceleration is proportional to displacement and acceleration is always directed towards a fixed point the motion of simple pendulum is “simple harmonic”.

Time period of simple pendulum :

Problem:
In simple pendulum T = 2π\(\sqrt{\frac{l}{g}}\) or l = \(\frac{gt^2}{4\pi^2}\)
For seconds pendulum T = 2s ⇒ t² = 4
∴ l = \(\sqrt{\frac{9.8\times4}{4\pi^2}}\) = 1 m (∴ π² nearly 9.8)

Question 14.
The period of a simple pendulum is found to increase by 50% when the length of the pendulum is increased by 0.6 m. Calculate the initial length and the initial period of oscillation at a place where g = 9.8 m/s².
Solution:
a) Increase in length of pendulum = 0.6m ;
Increase in time period = 50% = 1.5T
Let original length of pendulum = 1
Original time period = T; g = 9.8 m/s².
For 1st case 9.8 = π² \(\frac{1}{T^2}\) → 1 ;
For 2nd case l₁ = (l + 0.6), T₁ = 1.5 T

But l₁ = l + 0.6 ;
∴ l + 0.6 = 2.25l ⇒ 0.6 = 1.25l
∴ Length of pendulum l = \(\frac{0.6}{1.25}\) = 0.48 m
b) Time period T = 2π\(\sqrt{\frac{l}{g}}\)
= 2 × 3.142\(\sqrt{\frac{0.48}{9.8}}\) = 6.284 × 0.2213
= 1.391 sec.

Question 15.
A clock regulated by a seconds pendulum keeps correct time. During summer the length of the pendulum increases to 1.02m. How much will the clock gain or lose in one day?
Solution:
Time period of seconds pendulum,
T = 2 sec
Length of seconds pendulum,
L = gT² / 4π² = 0. 9927 m
Length of seconds pendulum during summer = 1.02 m
∴ Error in length, ∆l = 1.02 – 1 = 0.0273
In pendulum T × √l. From principles of error

Question 16.
The time period of a body suspended from a spring is T. What will be the new time period, if the spring is cut into two equal parts and (i) the mass is suspended from one part? (ii) the mass is suspended simultaneously from both the parts?
Solution:
Time period of spring, T = 2π\(\sqrt{\frac{m}{K}}\)
When a spring is cut into two equal parts force constant of each part K₁ = 2K

ii) When mass is suspended simultaneously from two parts ⇒ they are connected in parallel. For springs in parallel K_p = K₁ – K₂ = 4K

Question 16.
What is the length of a seconds pendulum on the earth? [AP Mar. ’17, ’16; June ’15; TS Mar. ’17]
Solution: