Oscillations

Notes Exam Tips Practice Qs AI Tutor Objectives

Learning Objectives

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Understand the fundamental concepts of oscillatory motion.
Describe the characteristics of periodic and oscillatory motions.
Explain simple harmonic motion (SHM) and its properties.
Analyze the relationship between velocity, acceleration, and displacement in SHM.
Apply the force law for simple harmonic motion in problem-solving.
Calculate energy in simple harmonic motion and understand its conservation.
Investigate the behavior of a simple pendulum and its approximation to SHM.
Explore the applications of oscillatory motion in real-world scenarios.

Revision Notes & Summary

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Chapter Thirteen: Oscillations

13.1 Introduction

Various kinds of motions in daily life include:
- Rectilinear motion
- Motion of a projectile
- Uniform circular motion
- Orbital motion of planets
Periodic Motion: Repetitive motion after a certain interval of time.
Oscillatory Motion: Repetitive to and fro motion about a mean position.
- Examples include:
  - Rocking in a cradle
  - Swinging on a swing
  - Pendulum of a wall clock
  - Boat tossing in a river
  - Piston in a steam engine
Key Concepts: Period, frequency, displacement, amplitude, and phase are fundamental to understanding oscillatory motion.

13.2 Periodic and Oscillatory Motions

Periodic Motion: Motion that repeats at regular intervals.
Oscillatory Motion: A type of periodic motion where the body oscillates about an equilibrium position.
Examples:
- Ball in a bowl oscillating when displaced.
- Bouncing ball off the ground.
Difference: All oscillatory motions are periodic, but not all periodic motions are oscillatory (e.g., circular motion).

13.3 Simple Harmonic Motion (SHM)

Definition: A type of oscillatory motion where the restoring force is proportional to the displacement from the mean position.
Mathematical Representation:
- Displacement:
  
  $x(t) = A ext{cos}(wt + \Phi)$
- Where:
  - A = Amplitude
  - w = Angular frequency
  - \Phi = Phase constant
Velocity and Acceleration:
- Velocity:
  
  $v(t) = -wA ext{sin}(wt + \Phi)$
- Acceleration:
  
  $a(t) = -w^2A ext{cos}(wt + \Phi)$

13.4 Force Law for Simple Harmonic Motion

Force:

$F = -kx$
- Where k is the force constant.

13.5 Energy in Simple Harmonic Motion

Kinetic Energy (K):

$K = \frac{1}{2} mv^2$
Potential Energy (U):

$U = \frac{1}{2} kx^2$
Total Mechanical Energy (E):

$E = K + U$ remains constant.

13.6 The Simple Pendulum

Approximation: The motion of a simple pendulum is approximately simple harmonic for small angular displacements.
Period of Oscillation:

$T = 2\pi \sqrt{\frac{L}{g}}$
- Where L is the length of the pendulum and g is the acceleration due to gravity.

13.7 Key Points to Ponder

Period T is the least time for motion to repeat.
Not all periodic motions are simple harmonic.
Circular motion can arise from various forces.
Initial conditions determine the motion in SHM.
Damped SHM is approximately simple harmonic for short time intervals.
The period of SHM does not depend on amplitude or energy.
A combination of SHM can be periodic only under certain conditions.

Exercises

Identify examples of periodic and simple harmonic motion.
Analyze graphs of motion to determine periodicity.
Solve problems related to SHM and its characteristics.

Exam Tips & Common Mistakes

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Common Mistakes and Exam Tips in Oscillations

Common Pitfalls

Confusing Periodic and Oscillatory Motion: Not every periodic motion is oscillatory. For example, circular motion is periodic but not oscillatory.
Misunderstanding Simple Harmonic Motion (SHM): Only periodic motion governed by the force law F = -kx is classified as SHM.
Ignoring Damping Effects: In real-world scenarios, oscillations eventually come to rest due to damping. Students often forget to consider this in their calculations.
Incorrect Application of Formulas: Ensure that the correct formulas for period and frequency are used, particularly in SHM where T = 2π√(m/k).

Exam Tips

Understand Key Concepts: Make sure to grasp fundamental concepts like period, frequency, amplitude, and phase. These are crucial for solving problems related to oscillations.
Practice with Graphs: Familiarize yourself with graphs of oscillatory motion. Being able to interpret these can help in identifying periodic behavior.
Use Dimensional Analysis: When deriving formulas, use dimensional analysis to check the consistency of units, especially for period and frequency.
Review Energy Concepts: Remember that in SHM, both kinetic and potential energies are periodic functions, and the total mechanical energy remains constant.
Check Initial Conditions: When solving problems, pay attention to initial conditions as they can significantly affect the outcome of SHM calculations.

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Practice Test – MCQs, True/False

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Multiple Choice Questions

2\pi \sqrt{\frac{h}{g}}

\pi \sqrt{\frac{h}{g}}

4\pi \sqrt{\frac{h}{g}}

\sqrt{\frac{h}{g}}

Correct Answer: A

Solution:

Given

\rho = \frac{1}{2}\rho_1

, the period

T = 2\pi \sqrt{\frac{h}{g(\rho_1 - \rho)}} = 2\pi \sqrt{\frac{h}{g(\rho_1 - \frac{1}{2}\rho_1)}} = 2\pi \sqrt{\frac{h}{g \cdot \frac{1}{2}\rho_1}} = 2\pi \sqrt{\frac{h}{g}}

1 second

2 seconds

0.5 seconds

4 seconds

Correct Answer: A

Solution:

The period

T

is given by

T = \frac{2\pi}{\omega}

. Substituting

\omega = 2\pi

, we get

T = \frac{2\pi}{2\pi} = 1

second.

Non-periodic motion

Periodic but not simple harmonic motion

Simple harmonic motion

Damped harmonic motion

Correct Answer: C

Solution:

When the suction pump is removed, the mercury column in the U-tube executes simple harmonic motion due to the restoring force provided by the pressure difference and gravity.

A car moving at constant speed on a straight road

A pendulum swinging with small angles

A ball rolling down a hill

A satellite orbiting the Earth

Correct Answer: B

Solution:

A pendulum swinging with small angles is an example of simple harmonic motion because its motion can be approximated by SHM equations.

10 cm/s

20 cm/s

31.4 cm/s

62.8 cm/s

Correct Answer: C

Solution:

The maximum velocity

V_{max}

in SHM is given by

V_{max} = \omega A

, where

A

is the amplitude. Here,

\omega = 2\pi

rad/s and

A = 5

cm. Thus,

V_{max} = 2\pi \times 5 = 31.4

cm/s.

2 rad/s

2\pi rad/s

\pi rad/s

4\pi rad/s

Correct Answer: B

Solution:

The displacement function is given by

x(t) = 5 \cos(2\pi t + \frac{\pi}{4})

. The general form of SHM is

x(t) = A \cos(\omega t + \Phi)

, where

\omega

is the angular frequency. Comparing the given equation with the standard form, we find that

\omega = 2\pi

rad/s.

T = \frac{1}{\omega}

T = 2\pi \omega

T = \frac{2\pi}{\omega}

T = \omega^2

Correct Answer: C

Solution:

The period

T

of a simple harmonic motion is related to the angular frequency

\omega

T = \frac{2\pi}{\omega}

The motion is non-periodic.

The force is proportional to the displacement and directed towards the mean position.

The velocity is constant.

The acceleration is zero.

Correct Answer: B

Solution:

In SHM, the force acting on the particle is proportional to the displacement and is always directed towards the mean position, which is the restoring force.

T = 2\pi \sqrt{\frac{\rho_c}{\rho_l g}}

T = 2\pi \sqrt{\frac{h \rho_c}{\rho_l g}}

T = 2\pi \sqrt{\frac{\rho_l}{\rho_c g}}

T = 2\pi \sqrt{\frac{h \rho_l}{\rho_c g}}

Correct Answer: B

Solution:

The period

T

for the oscillation of the cork is given by

T = 2\pi \sqrt{\frac{h \rho_c}{\rho_l g}}

, where

h

is the height of the cork and

g

is the acceleration due to gravity.

A car moving on a straight road

A pendulum swinging back and forth

A satellite orbiting the Earth

A person walking in a park

Correct Answer: B

Solution:

Oscillatory motion is characterized by repetitive to and fro movement about a mean position, such as a pendulum swinging back and forth.

It remains the same.

It doubles.

It halves.

It quadruples.

Correct Answer: B

Solution:

The period of a simple pendulum is given by

T = 2\pi \sqrt{\frac{L}{g}}

. If

g

is reduced to

\frac{g}{4}

, then the new period

T' = 2\pi \sqrt{\frac{L}{g/4}} = 2\pi \sqrt{\frac{4L}{g}} = 2 \times 2\pi \sqrt{\frac{L}{g}} = 2T

. Thus, the period doubles.

Newton

Hertz

Joule

Pascal

Correct Answer: B

Solution:

The unit of frequency in the SI system is Hertz (Hz), which is equivalent to one oscillation per second.

10 cm/s

50 cm/s

100 cm/s

150 cm/s

Correct Answer: C

Solution:

The maximum velocity in SHM is given by

v_{max} = wA

, where

w

is the angular frequency and

A

is the amplitude. Substituting

w = 5

rad/s and

A = 10

cm, we get

v_{max} = 5 \times 10 = 50

cm/s.

0.23 s

0.36 s

0.45 s

0.52 s

Correct Answer: B

Solution:

The period of oscillation for a spring-mass system is given by

T = 2\pi \sqrt{\frac{m}{k}}

. Here,

m = 2

kg and

k = 1500

N/m. So,

T = 2\pi \sqrt{\frac{2}{1500}} \approx 0.36

0

\frac{\pi}{2}

\pi

\frac{3\pi}{2}

Correct Answer: A

Solution:

For the particle to be at its maximum positive displacement at

t = 0

x(0) = A \cos(\phi) = A

. This implies

\cos(\phi) = 1

, so

\phi = 0

\sqrt{\frac{k}{m}}

\frac{k}{m}

\sqrt{\frac{m}{k}}

\frac{m}{k}

Correct Answer: A

Solution:

The angular frequency

\omega

of a simple harmonic oscillator is given by

\omega = \sqrt{\frac{k}{m}}

\Phi = 0

\Phi = \frac{\pi}{2}

\Phi = \pi

\Phi = \frac{3\pi}{2}

Correct Answer: B

Solution:

If the displacement is zero at

t = 0

, the phase constant

\Phi

must be

\frac{\pi}{2}

for the function

x(t) = A \cos(\omega t + \Phi)

to satisfy this condition.

0.2 s

5 s

2 s

0.5 s

Correct Answer: A

Solution:

The period

T

is the reciprocal of the frequency

v

. Thus,

T = \frac{1}{v} = \frac{1}{5} = 0.2

T/2

T/\sqrt{2}

T/\sqrt{4}

T/4

Correct Answer: A

Solution:

The period of oscillation for a mass-spring system is

T = 2\pi \sqrt{\frac{m}{k}}

. If

k

is doubled and

m

is halved, the new period

T' = 2\pi \sqrt{\frac{m/2}{2k}} = T/2

T = v

T = \frac{1}{v}

T = 2\pi v

T = v^2

Correct Answer: B

Solution:

The period

T

is the reciprocal of the frequency

v

, i.e.,

T = \frac{1}{v}

1.5 seconds

3.5 seconds

7.5 seconds

9.5 seconds

Correct Answer: C

Solution:

The period of a pendulum is given by

T = 2\pi \sqrt{\frac{l}{g}}

. The ratio of periods is

\frac{T_{moon}}{T_{earth}} = \sqrt{\frac{g_{earth}}{g_{moon}}}

. Substituting the values,

T_{moon} = T_{earth} \times \sqrt{\frac{9.8}{1.7}} = 3.5 \times \sqrt{5.76} = 7.5

seconds.

2.83 s

2 s

4 s

1.41 s

Correct Answer: A

Solution:

The period of a simple pendulum is given by

T = 2\pi \sqrt{\frac{l}{g}}

. On Earth,

T_{Earth} = 2\pi \sqrt{\frac{1}{9.8}}

. On the planet,

T_{planet} = 2\pi \sqrt{\frac{1}{4.9}} = 2\pi \sqrt{\frac{2}{9.8}} = \sqrt{2} \times T_{Earth}

. Therefore, the period on the planet is approximately 2.83 s.

T = \frac{1}{\omega}

T = 2\pi \omega

T = \frac{2\pi}{\omega}

T = \omega

Correct Answer: C

Solution:

The period

T

of simple harmonic motion is related to the angular frequency

\omega

by the formula

T = \frac{2\pi}{\omega}

10 m/s

3.33 m/s

6.67 m/s

1.0 m/s

Correct Answer: B

Solution:

The maximum speed

v_{max}

in SHM is given by

v_{max} = \omega A

, where

A

is the amplitude (half the stroke).

12 m/s

6 m/s

4 m/s

3 m/s

Correct Answer: A

Solution:

The maximum speed of a particle in simple harmonic motion is given by

v_{\text{max}} = \omega A

, where

\omega

is the angular frequency and

A

is the amplitude. Here,

\omega = 3\pi

rad/s and

A = 4

m. Thus,

v_{\text{max}} = 3\pi \times 4 = 12

m/s.

4 seconds

2√2 seconds

8 seconds

2 seconds

Correct Answer: A

Solution:

The period of a simple pendulum is given by

T = 2\pi \sqrt{\frac{L}{g}}

. If the length

L

is quadrupled, the new period

T' = 2\pi \sqrt{\frac{4L}{g}} = 2\times 2\pi \sqrt{\frac{L}{g}} = 2T = 4

seconds.

6 seconds

3 seconds

12 seconds

1.5 seconds

Correct Answer: A

Solution:

The period of a spring-mass system is given by

T = 2\pi \sqrt{\frac{m}{k}}

. If the spring constant

k

is reduced to one-fourth, the new period

T' = 2\pi \sqrt{\frac{m}{k/4}} = 2\pi \sqrt{4 \frac{m}{k}} = 2T

. Thus,

T' = 2 \times 3 = 6

seconds.

T/\sqrt{2}

T

\sqrt{2}T

2T

Correct Answer: C

Solution:

The period of a simple pendulum is given by

T = 2\pi \sqrt{\frac{L}{g}}

. If the gravitational acceleration

g

is halved, the new period

T' = 2\pi \sqrt{\frac{L}{g/2}} = 2\pi \sqrt{\frac{2L}{g}} = \sqrt{2} \cdot 2\pi \sqrt{\frac{L}{g}} = \sqrt{2}T

\sin(\omega t) - \cos(\omega t)

\sin^3(\omega t)

\exp(-\omega^2 t^2)

\cos(\omega t) + \cos(3\omega t) + \cos(5\omega t)

Correct Answer: C

Solution:

The function

\exp(-\omega^2 t^2)

is non-periodic as it does not repeat over time. It represents an exponentially decaying function.

f/\sqrt{2}

f

\sqrt{2}f

2f

Correct Answer: C

Solution:

The frequency of a mass-spring system is given by

f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

. If the spring constant

k

is doubled, the new frequency

f' = \frac{1}{2\pi} \sqrt{\frac{2k}{m}} = \sqrt{2} \cdot \frac{1}{2\pi} \sqrt{\frac{k}{m}} = \sqrt{2}f

v(t) = -\omega A \sin(\omega t + \Phi)

v(t) = \omega A \cos(\omega t + \Phi)

v(t) = A \sin(\omega t + \Phi)

v(t) = -A \cos(\omega t + \Phi)

Correct Answer: A

Solution:

The velocity of a particle in simple harmonic motion is given by the derivative of the displacement function:

v(t) = -\omega A \sin(\omega t + \Phi)

2 seconds

4.9 seconds

1.2 seconds

3.5 seconds

Correct Answer: B

Solution:

The time period of a pendulum is given by

T = 2\pi \sqrt{\frac{l}{g}}

. Since the gravity on the Moon is less, the time period will increase.

0

\frac{\pi}{2}

\pi

\frac{3\pi}{2}

Correct Answer: A

Solution:

If the initial position is

A

and the initial velocity is zero, the cosine function is at its maximum, which occurs when

\Phi = 0

a(t) = -\omega^2 A \cos(\omega t + \Phi)

a(t) = \omega^2 A \sin(\omega t + \Phi)

a(t) = -\omega A \sin(\omega t + \Phi)

a(t) = \omega A \cos(\omega t + \Phi)

Correct Answer: A

Solution:

The acceleration of a particle in SHM is given by

a(t) = -\omega^2 x(t) = -\omega^2 A \cos(\omega t + \Phi)

5 m

2 m

10 m

1 m

Correct Answer: A

Solution:

The amplitude of simple harmonic motion is the maximum displacement from the mean position, which is given by the coefficient of the cosine function. Therefore, the amplitude is 5 m.

\omega t + \Phi

A \cos(\omega t + \Phi)

\Phi

\omega t

Correct Answer: A

Solution:

The phase of the motion in simple harmonic motion is given by

\omega t + \Phi

T/3

T/9

Correct Answer: A

Solution:

The period of a mass-spring system is given by

T = 2\pi \sqrt{\frac{m}{k}}

. If the spring constant

k

is increased by a factor of 9, the new period

T' = 2\pi \sqrt{\frac{m}{9k}} = \frac{1}{3}T

. Thus, the new period is

T/3

2\pi\sqrt{h/g}

2\pi\sqrt{\rho/\rho_1 g}

2\pi\sqrt{\rho_1/\rho g}

2\pi\sqrt{h\rho/\rho_1 g}

Correct Answer: D

Solution:

The period of oscillation for a floating object in simple harmonic motion is given by T = 2\pi\sqrt{h\rho/\rho_1 g}, where h is the height of the cork,

\rho

is the density of the cork, and

\rho_1

is the density of the liquid.

1 cm

\sqrt{2}

\frac{1}{\omega}

2 cm

Correct Answer: B

Solution:

Using the initial conditions,

x(0) = A \cos(\Phi) = 1

and

v(0) = -A \omega \sin(\Phi) = \omega

. Solving these equations gives

A = \sqrt{2}

cm.

0.75 s

1.06 s

1.5 s

2.12 s

Correct Answer: C

Solution:

The period of oscillation of a liquid column in a U-tube is independent of the density of the liquid. Therefore, even if the density of mercury is doubled, the period remains the same, i.e., 1.5 s.

7.3 s

8.5 s

9.2 s

10.5 s

Correct Answer: A

Solution:

The period of a simple pendulum is given by

T = 2\pi \sqrt{\frac{l}{g}}

. On Earth,

T_{Earth} = 2\pi \sqrt{\frac{l}{9.8}} = 3.5

s. On the Moon,

T_{Moon} = 2\pi \sqrt{\frac{l}{1.7}}

. The ratio

\frac{T_{Moon}}{T_{Earth}} = \sqrt{\frac{9.8}{1.7}}

. Solving gives

T_{Moon} \approx 7.3

2\pi \sqrt{\frac{h}{g}}

2\pi \sqrt{\frac{\rho_c h}{\rho_l g}}

2\pi \sqrt{\frac{\rho_l h}{\rho_c g}}

2\pi \sqrt{\frac{g}{h}}

Correct Answer: B

Solution:

The period of oscillation for the cork is given by

T = 2\pi \sqrt{\frac{\rho_c h}{\rho_l g}}

. This is derived from the balance of forces and the buoyant force acting on the cork.

The force is always directed away from the mean position.

The force is proportional to the square of the displacement.

The force is proportional to the displacement and directed towards the mean position.

The force is constant throughout the motion.

Correct Answer: C

Solution:

In simple harmonic motion, the force is proportional to the displacement and is always directed towards the mean position.

4 cm

2 cm

8 cm

0 cm

Correct Answer: A

Solution:

The maximum displacement from the mean position in simple harmonic motion is equal to the amplitude, which is 4 cm.

5 Hz

2.5 Hz

10 Hz

3.54 Hz

Correct Answer: D

Solution:

The frequency of oscillation for a mass-spring system is given by

f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

. If the mass is doubled, the new frequency becomes

f' = \frac{1}{2\pi} \sqrt{\frac{k}{2m}} = \frac{f}{\sqrt{2}} = \frac{5}{\sqrt{2}} \approx 3.54 \text{ Hz}

It remains the same.

It doubles.

It halves.

It quadruples.

Correct Answer: C

Solution:

The angular frequency

\omega

of a simple harmonic oscillator is given by

\omega = \sqrt{\frac{k}{m}}

. If the mass

m

is increased by a factor of 4, the new angular frequency

\omega' = \sqrt{\frac{k}{4m}} = \frac{1}{2} \sqrt{\frac{k}{m}} = \frac{1}{2} \omega

. Thus, the angular frequency halves.

0.25 seconds

4 seconds

0.5 seconds

1 second

Correct Answer: A

Solution:

The period

T

is the time for one complete oscillation. Since 20 oscillations take 5 seconds, the period is

T = \frac{5}{20} = 0.25

seconds.

The velocity of the particle

The displacement of the particle

Solution:

The maximum acceleration

a_{max}

in SHM is given by

a_{max} = \omega^2 A

. Substituting

\omega = 5

rad/s and

A = 4

cm, we get

a_{max} = (5)^2 \times 4 = 100

cm/s².

Correct Answer: A

Solution:

The amplitude of SHM is the coefficient of the cosine function in the displacement equation, which is 4.

0.2 seconds

5 seconds

0.5 seconds

2 seconds

Correct Answer: A

Solution:

The period

T

is the time for one complete oscillation. If 50 oscillations take 10 seconds, then

T = \frac{10}{50} = 0.2

seconds.

\pi/3

\pi/6

\pi/2

0

Correct Answer: A

Solution:

The phase of the motion is given by

(3t + \pi/3)

. At

t = 0

, the phase is

3 \times 0 + \pi/3 = \pi/3

A car moving on a straight road

The Earth revolving around the Sun

A stone falling freely under gravity

A person walking in a park

Correct Answer: B

Solution:

The Earth revolving around the Sun is a periodic motion because it repeats after a fixed interval of time.

0

\frac{\pi}{2}

\pi

2\pi

Correct Answer: C

Solution:

In simple harmonic motion, the acceleration is out of phase with the displacement by

\pi

radians.

T = 2\pi \sqrt{\frac{\rho_c}{\rho_l g}}

T = 2\pi \sqrt{\frac{\rho_l}{\rho_c g}}

T = 2\pi \sqrt{\frac{\rho_c}{\rho_l}}

T = 2\pi \sqrt{\frac{\rho_l}{\rho_c}}

Correct Answer: A

Solution:

The period of oscillation for a floating object is given by

T = 2\pi \sqrt{\frac{h \rho_c}{\rho_l g}}

, where

h

is the height of the cork. This simplifies to

T = 2\pi \sqrt{\frac{\rho_c}{\rho_l g}}

when considering the effective height.

0

6

-6

12

Correct Answer: C

Solution:

The velocity

v(t)

is given by

v(t) = -\omega A \sin(\omega t + \phi)

. Here,

A = 4

\omega = 3

, and

\phi = \frac{\pi}{6}

. At

t = 0

v(0) = -3 \times 4 \times \sin(\frac{\pi}{6}) = -12 \times \frac{1}{2} = -6

x(t) = \sin^3(\omega t)

x(t) = \cos(\omega t)

x(t) = e^{-\omega t^2}

x(t) = 1 + \omega t

Correct Answer: B

Solution:

Simple harmonic motion is characterized by sinusoidal functions like

\cos(\omega t)

\sin(\omega t)

. Therefore,

x(t) = \cos(\omega t)

represents SHM.

\frac{1}{2} k x

k x^2

\frac{1}{2} k x^2

k x

Correct Answer: C

Solution:

The potential energy stored in a compressed or stretched spring is given by

U = \frac{1}{2} k x^2

1.18 m/s²

2.37 m/s²

3.55 m/s²

4.73 m/s²

Correct Answer: B

Solution:

The acceleration in simple harmonic motion is given by

a = -\omega^2 x

, where

\omega = \frac{2\pi}{T}

and

x

is the displacement. Here,

T = 2

s, so

\omega = \pi

rad/s. At

x = 0.3

a = -\pi^2 \times 0.3 \approx -2.37

m/s².

Velocity: Positive, Acceleration: Negative, Force: Negative

Velocity: Negative, Acceleration: Positive, Force: Positive

Velocity: Positive, Acceleration: Positive, Force: Positive

Velocity: Negative, Acceleration: Negative, Force: Negative

Correct Answer: B

Solution:

At the midpoint going towards A, the velocity is negative (since it is moving towards A), the acceleration is positive (since it is directed towards the center), and the force is also positive (since it is a restoring force directed towards the center).

T = 2\pi \sqrt{\frac{m}{k}}

T = 2\pi \sqrt{\frac{k}{m}}

T = \frac{1}{2\pi} \sqrt{\frac{m}{k}}

T = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

Correct Answer: A

Solution:

The period of oscillation for a mass-spring system is given by

T = 2\pi \sqrt{\frac{m}{k}}

1.15 s

1.63 s

2.58 s

3.14 s

Correct Answer: A

Solution:

The period of a simple pendulum is given by

T = 2\pi \sqrt{\frac{L}{g}}

. Here,

L = 2

m and

g = 15

m/s². Thus,

T = 2\pi \sqrt{\frac{2}{15}} \approx 1.15

-kx

kx

\frac{k}{x}

\frac{x}{k}

Correct Answer: A

Solution:

The force acting on a spring is given by Hooke's law,

F = -kx

, where

k

is the spring constant and

x

is the displacement.

1 Hz

1.41 Hz

2 Hz

0.5 Hz

Correct Answer: B

Solution:

The frequency of oscillation for a spring-mass system is given by

f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

. If the mass is doubled, the new frequency

f'

becomes

f' = \frac{1}{2\pi} \sqrt{\frac{k}{2m}} = \frac{f}{\sqrt{2}}

. Thus, the new frequency is

\frac{2}{\sqrt{2}} = 1.41

Hz.

k/2

2k

k

k/4

Correct Answer: A

Solution:

For two springs in series, the effective spring constant

k_{eff}

is given by

\frac{1}{k_{eff}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k}

, thus

k_{eff} = k/2

10 Hz

3.2 Hz

5.6 Hz

2.0 Hz

Correct Answer: B

Solution:

The frequency of oscillation for a spring-mass system is given by

f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

. Substituting the given values,

f = \frac{1}{2\pi} \sqrt{\frac{1200}{3}} \approx 3.2 \text{ Hz}

1 second

2 seconds

0.5 seconds

4 seconds

Correct Answer: A

Solution:

The period of a simple pendulum is given by

T = 2\pi \sqrt{\frac{L}{g}}

. If the acceleration due to gravity is increased by a factor of 4, the new period will be

T = 2\pi \sqrt{\frac{L}{4g}} = \frac{1}{2} \times 2 = 1 \text{ second}

\pi/3

\pi/6

\pi/4

\pi/2

Correct Answer: A

Solution:

The initial phase angle

\Phi

in the equation

x(t) = A \cos(\omega t + \Phi)

is the constant term inside the cosine function. Here,

\Phi = \pi/3

3.54 cm

2.5 cm

5 cm

0 cm

Correct Answer: B

Solution:

In SHM, the kinetic energy (K) and potential energy (U) are equal when the displacement x is such that K = U. Since the total energy E = K + U is constant and given by E = 1/2 k A^2, where A is the amplitude, we have K = U = 1/4 k A^2. The displacement x where this occurs is given by x = A/√2. Substituting A = 5 cm, we get x = 5/√2 = 3.54 cm.

It remains the same.

It doubles.

It quadruples.

It halves.

Correct Answer: C

Solution:

The maximum kinetic energy in SHM is given by

K_{\text{max}} = \frac{1}{2}kA^2

. If the amplitude

A

is doubled,

A' = 2A

, then

K'_{\text{max}} = \frac{1}{2}k(2A)^2 = 4 \times \frac{1}{2}kA^2 = 4K_{\text{max}}

. Thus, the maximum kinetic energy quadruples.

0.5 Hz

2 Hz

1 Hz

0.25 Hz

Correct Answer: A

Solution:

The frequency

f

is the reciprocal of the period

T

. Therefore,

f = \frac{1}{T} = \frac{1}{2} = 0.5

Hz.

Non-periodic motion

Damped harmonic motion

Simple harmonic motion

Uniform circular motion

Correct Answer: C

Solution:

When the suction pump is removed, the mercury column in the U-tube executes simple harmonic motion due to the restoring force provided by gravity acting on the displaced mercury.

\sin^3(\omega t)

\cos(\omega t) + \cos(3\omega t) + \cos(5\omega t)

\sin(\omega t) - \cos(\omega t)

\cos(\omega t)

Correct Answer: D

Solution:

Simple harmonic motion is represented by functions of the form

A \cos(\omega t + \phi)

A \sin(\omega t + \phi)

. Among the given options, only

\cos(\omega t)

fits this form.

12 m/s²

36 m/s²

48 m/s²

24 m/s²

Correct Answer: B

Solution:

Solution:

The angular frequency

\omega

of a simple harmonic oscillator is given by

\omega = \sqrt{\frac{k}{m}}

, where

k

is the force constant and

m

is the mass.

Correct Answer: True

Solution:

The displacement in simple harmonic motion can be expressed as

x(t) = A \cos(\omega t + \Phi)

x(t) = B \sin(\omega t + \alpha)

Correct Answer: True

Solution:

Both kinetic energy and potential energy in simple harmonic motion vary periodically with time, reaching maximum values at different points in the cycle.

Correct Answer: True

Solution:

In simple harmonic motion, the acceleration is always directed towards the mean position, acting as a restoring force.

Correct Answer: False

Solution:

In simple harmonic motion, the kinetic energy varies with time, being maximum at the equilibrium position and zero at the extreme positions.

Correct Answer: False

Solution:

The force in simple harmonic motion is linearly proportional to the displacement, given by

F = -kx

, not proportional to the square of the displacement.

Correct Answer: True

Solution:

The period of a mass-spring system executing simple harmonic motion is determined by the mass and the spring constant, and it does not depend on the amplitude of oscillation.

Correct Answer: False

Solution:

In simple harmonic motion, the velocity and acceleration are not in phase. The velocity leads the displacement by a phase of

\pi/2

, and the acceleration leads the displacement by a phase of

\pi

Correct Answer: True

Solution:

Simple harmonic motion is defined by a restoring force that is proportional to the negative of the displacement, as given by the equation

F = -kx

Correct Answer: True

Solution:

Oscillatory motion is characterized by the repetitive to and fro movement about a mean position, which is exactly how a pendulum in a wall clock moves.

Correct Answer: True

Solution:

Frequency is defined as the number of oscillations or cycles completed per unit time, typically measured in hertz (Hz).

Correct Answer: True

Solution:

Oscillatory motion is characterized by a repetitive to and fro movement about a mean position, which is exemplified by the motion of a pendulum.

Correct Answer: True

Solution:

The period

T

Correct Answer: True

Solution:

The displacement in simple harmonic motion can be expressed as

x = A \cos(\omega t + \phi)

Solution:

This equation represents the displacement of a particle in simple harmonic motion, where

A

is the amplitude,

\omega

is the angular frequency, and

Solution:

Not all periodic motions are simple harmonic. A periodic motion is only simple harmonic if it follows the force law

F = -kx

. Other periodic motions may not satisfy this condition.

Correct Answer: True

Solution:

In simple harmonic motion, the speed is maximum at the mean position and zero at the extreme positions.

Correct Answer: False

Solution:

The period of simple harmonic motion does not depend on the amplitude; it is determined by the properties of the system such as mass and spring constant.

Correct Answer: True

Solution:

The displacement in simple harmonic motion is described by the equation

X(t) = A \cos(\omega t + \Phi)

, where

A

is the amplitude,

\omega

is the angular frequency, and

Solution:

In simple harmonic motion, if no friction is present, the mechanical energy, which is the sum of kinetic and potential energy, remains constant.

Correct Answer: False

Solution:

Not every periodic motion is simple harmonic. Only periodic motion governed by the force law

F = -kx

is simple harmonic.

Oscillations

Learning Objectives

Revision Notes & Summary

Chapter Thirteen: Oscillations

13.1 Introduction

13.2 Periodic and Oscillatory Motions

13.3 Simple Harmonic Motion (SHM)

13.4 Force Law for Simple Harmonic Motion

13.5 Energy in Simple Harmonic Motion

13.6 The Simple Pendulum

13.7 Key Points to Ponder

Exercises

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips in Oscillations

Common Pitfalls

Exam Tips

AI Learning Assistant

Practice Test – MCQs, True/False

Multiple Choice Questions

Correct Answer: A

Q2.If a particle executing simple harmonic motion has an angular frequency ω=2π\omega = 2 \piω=2π rad/s, what is its period TTT?

Correct Answer: A

Q3.In a U-tube filled with mercury, a small pressure difference causes the mercury column to oscillate. Which of the following best describes the motion of the mercury column?

Correct Answer: C

Q4.Which of the following motions is an example of simple harmonic motion?

Correct Answer: B

Q5.Consider a particle executing simple harmonic motion (SHM) with an angular frequency of ω=2π\omega = 2\piω=2π rad/s. If the amplitude of the motion is 5 cm, what is the maximum velocity of the particle?

Correct Answer: C

Q6.A particle is executing simple harmonic motion (SHM) with a displacement given by x(t)=5cos⁡(2πt+π4)x(t) = 5 \cos(2\pi t + \frac{\pi}{4})x(t)=5cos(2πt+4π​). What is the angular frequency of this motion?

Correct Answer: B

Q7.What is the relationship between the period TTT and the angular frequency ω\omegaω of a simple harmonic motion?

Correct Answer: C

Q8.Which of the following is a characteristic of simple harmonic motion (SHM)?

Correct Answer: B

Q9.A cylindrical piece of cork of density ρc\rho_cρc​ and base area AAA floats in a liquid of density ρl\rho_lρl​. If the cork is slightly depressed and released, it oscillates with a period TTT. Which expression correctly describes TTT?

Correct Answer: B

Q10.Which of the following motions is an example of oscillatory motion?

Correct Answer: B

Q11.A simple pendulum has a length LLL and swings with a small amplitude on Earth. If the pendulum is taken to a planet where the gravitational acceleration is one-fourth that of Earth's, what happens to the period of the pendulum?

Correct Answer: B

Q12.What is the unit of frequency in the SI system?

Correct Answer: B

Q13.A particle is executing simple harmonic motion (SHM) with an amplitude of 10 cm and an angular frequency of 5 rad/s. What is the maximum velocity of the particle?

Correct Answer: C

Q14.A spring-mass system with a spring constant of 1500 N/m and a mass of 2 kg is oscillating. What is the period of oscillation?

Correct Answer: B

Q15.A particle is executing simple harmonic motion (SHM) with a displacement given by x(t)=Acos⁡(ωt+ϕ)x(t) = A \cos(\omega t + \phi)x(t)=Acos(ωt+ϕ). If the particle is at its maximum positive displacement at t=0t = 0t=0, what is the initial phase constant ϕ\phiϕ?

Correct Answer: A

Q16.A spring with a spring constant kkk is attached to a mass mmm. If the mass is displaced and released, what is the angular frequency of the resulting simple harmonic motion?

Correct Answer: A

Q17.A particle is executing simple harmonic motion with an angular frequency ω\omegaω. If its displacement is zero at t=0t = 0t=0, what is the phase constant Φ\PhiΦ?

Correct Answer: B

Q18.What is the period of a simple harmonic motion if its frequency is 5 Hz?

Correct Answer: A

Q19.A mass-spring system oscillates with a period TTT. If the spring constant is doubled and the mass is halved, what will be the new period of oscillation?

Correct Answer: A

Q20.What is the relationship between the period TTT and frequency vvv of a periodic motion?

Correct Answer: B

Q21.What is the period of a simple pendulum on the moon if its period on Earth is 3.5 seconds? (Given: acceleration due to gravity on the moon is 1.7 m/s² and on Earth is 9.8 m/s²)

Correct Answer: C

Q22.Consider a simple pendulum with a length of 1 m on Earth where the acceleration due to gravity is 9.8 m/s². What would be the period of oscillation if the pendulum is taken to a planet where the acceleration due to gravity is 4.9 m/s²?

Correct Answer: A

Q23.If a particle is in simple harmonic motion with an angular frequency ω\omegaω, what is the relationship between its period TTT and ω\omegaω?

Correct Answer: C

Q24.What is the maximum speed of a piston moving with simple harmonic motion with an angular frequency of 200 rad/min and a stroke of 1.0 m?

Correct Answer: B

Q25.A particle is executing simple harmonic motion with a displacement given by x(t)=4cos⁡(3πt+π6)x(t) = 4 \cos(3\pi t + \frac{\pi}{6})x(t)=4cos(3πt+6π​). What is the maximum speed of the particle?

Correct Answer: A

Q26.A simple pendulum on Earth has a period of 2 seconds. If the length of the pendulum is quadrupled, what will be the new period of oscillation?

Correct Answer: A

Q27.A mass attached to a spring is oscillating with a period of 3 seconds. If the spring constant is reduced to one-fourth of its original value, what will be the new period of oscillation?

Correct Answer: A

Q28.A simple pendulum with length LLL and mass mmm swings with a small amplitude. If the pendulum is taken to a planet where the gravitational acceleration is half that of Earth, what will be the new period of oscillation?

Correct Answer: C

Q29.Which of the following functions represents a non-periodic motion?

Correct Answer: C

Q30.A mass-spring system oscillates with a frequency fff. If the spring constant is doubled while the mass remains constant, what will be the new frequency of oscillation?

Correct Answer: C

Q31.If a particle is executing simple harmonic motion with a displacement function x(t)=Acos⁡(ωt+Φ)x(t) = A \cos(\omega t + \Phi)x(t)=Acos(ωt+Φ), what is the expression for its velocity?

Correct Answer: A