Oscillations

Study Action Plan (Sorted by Difficulty)
Damped and Forced Oscillations
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Displacement, Amplitude, Phase and Angular Frequency
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Energy in SHM
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Force Law in SHM
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Identifying SHM from Functions and Graphs
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Initial Conditions in SHM
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Period and Frequency
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Periodic and Oscillatory Motions
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Projection of Circular Motion
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Simple Harmonic Motion (SHM)
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Simple Pendulum
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Spring-Mass System and Linear Oscillator
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Velocity and Acceleration in SHM
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Learning Objectives

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Learning Objectives

  • Understand the mathematical representation of Simple Harmonic Motion (SHM) using the equation X(t)=Acos⁡(ωt+Φ)X(t) = A \cos(\omega t + \Phi), where AA is the amplitude, ω\omega is the angular frequency, and Φ\Phi is the phase constant.
  • Calculate the energy variations in SHM, recognizing that the total mechanical energy remains constant and is given by E=12kA2E = \frac{1}{2} kA^2, where kk is the spring constant.
  • Apply the force law in SHM, expressed as F(t)=−kx(t)F(t) = -kx(t), to analyze the proportional relationship between force and displacement.
  • Derive the velocity and acceleration equations in SHM, V(t)=−ωAsin⁡(ωt+Φ)V(t) = -\omega A \sin(\omega t + \Phi) and a(t)=−ω2Acos⁡(ωt+Φ)a(t) = -\omega^2 A \cos(\omega t + \Phi), respectively, and understand their periodic nature.
  • Examine the motion of a simple pendulum and derive its period T=2πLgT = 2\pi\sqrt{\frac{L}{g}}, where LL is the length of the pendulum and gg is the acceleration due to gravity.
  • Relate the period TT and frequency vv of oscillatory motions using T=1vT = \frac{1}{v}, and understand the units of frequency in hertz (Hz).
  • Visualize SHM as the projection of uniform circular motion, linking angular motion to linear oscillations.
  • Differentiate between damped and forced oscillations, understanding the effects of dissipative forces and external periodic forces.
  • Distinguish between periodic and oscillatory motions, identifying that every oscillatory motion is periodic, but not every periodic motion is oscillatory.
  • Identify SHM from functions and graphs, determining periodicity and calculating the period from given functions.
  • Analyze the spring-mass system and linear oscillator, using the relation ω=km\omega = \sqrt{\frac{k}{m}} and the period T=2πmkT = 2\pi\sqrt{\frac{m}{k}}.
  • Determine initial conditions in SHM, such as amplitude and phase constant, from initial position and velocity.

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Revision Notes & Summary

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Chapter Notes

Simple Harmonic Motion (SHM)

  • Definition: SHM is characterized by the displacement of a particle from its equilibrium position described by X(t)=Acos⁡(ωt+Φ)X(t) = A \cos(\omega t + \Phi).
    • Amplitude (AA): Maximum displacement from the equilibrium position.
    • Angular Frequency (ω\omega): Related to the period and frequency of the motion.
    • Phase Constant (Φ\Phi): Initial phase of the motion.
  • Projection of Circular Motion: SHM can be visualized as the projection of uniform circular motion on a diameter of the circle.

Energy in SHM

  • Total Mechanical Energy: E=12kA2E = \frac{1}{2} kA^2, where kk is the spring constant.
  • Kinetic Energy (KK): Varies periodically, zero at maximum displacement.
  • Potential Energy (UU): Also varies periodically, maximum at maximum displacement.

Force Law in SHM

  • Restoring Force: F(t)=−kx(t)F(t) = -kx(t), where kk is the force constant.
  • Proportional to Displacement: Always directed towards the equilibrium position.

Velocity and Acceleration in SHM

  • Velocity (V(t)V(t)): V(t)=−ωAsin⁡(ωt+Φ)V(t) = -\omega A \sin(\omega t + \Phi).
  • Acceleration (a(t)a(t)): a(t)=−ω2Acos⁡(ωt+Φ)a(t) = -\omega^2 A \cos(\omega t + \Phi).

Simple Pendulum

  • Period (TT): T=2πLgT = 2\pi \sqrt{\frac{L}{g}}, where LL is the length of the pendulum and gg is the acceleration due to gravity.
  • Small Angular Displacements: Motion is approximately simple harmonic.

Period and Frequency

  • Period (TT): Time for one complete oscillation, T=1vT = \frac{1}{v}.
  • Frequency (vv): Number of oscillations per unit time, measured in hertz (Hz).

Damped and Forced Oscillations

  • Damping: Due to dissipative forces.
  • Forced Oscillations: Due to external periodic agency.

Periodic and Oscillatory Motions

  • Periodic Motion: Repeats itself at regular intervals.
  • Oscillatory Motion: To and fro motion about a mean position.
  • Distinction: Every oscillatory motion is periodic, but not every periodic motion is oscillatory.

Displacement, Amplitude, Phase, and Angular Frequency

  • Displacement Variable: Change with time of any physical property.
  • Representation: Using sine/cosine functions.

Identifying SHM from Functions and Graphs

  • Periodic Functions: Can be represented as a sum of sine and cosine functions.
  • Graph Analysis: Determine periodicity and SHM characteristics.

Spring-Mass System and Linear Oscillator

  • Block-Spring Oscillator: Angular frequency ω=km\omega = \sqrt{\frac{k}{m}}, period T=2πmkT = 2\pi \sqrt{\frac{m}{k}}.

Initial Conditions in SHM

  • Determining Parameters: Amplitude, phase constant, and equation of SHM from initial position and velocity.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

x(t)=Acos⁡(ωt+ϕ)x(t) = A \cos(\omega t + \phi)

B.

x(t)=Asin⁡(ωt+ϕ)x(t) = A \sin(\omega t + \phi)

C.

x(t)=Ae−ωtx(t) = A e^{-\omega t}

D.

x(t)=Atan⁡(ωt+ϕ)x(t) = A \tan(\omega t + \phi)
Correct Answer: A

Solution:

The displacement in SHM is given by x(t)=Acos⁡(ωt+ϕ)x(t) = A \cos(\omega t + \phi), where AA is the amplitude, ω\omega is the angular frequency, and ϕ\phi is the phase constant.

Chapter Concept:

Displacement, Amplitude, Phase and Angular Frequency

A.

The motion of a pendulum with small angular displacement.

B.

The rotation of the Earth about its axis.

C.

The motion of a ball bearing inside a smooth curved bowl.

D.

The general vibrations of a polyatomic molecule about its equilibrium position.
Correct Answer: A

Solution:

Simple harmonic motion is characterized by a restoring force proportional to displacement. The motion of a pendulum with small angular displacement fits this description.

Chapter Concept:

Periodic and Oscillatory Motions

A.

1 cm

B.

2\sqrt{2} cm

C.

22 cm

D.

1ω\frac{1}{\omega} cm
Correct Answer: B

Solution:

The amplitude AA can be found using the initial conditions. Given x(0)=1x(0) = 1 cm and v(0)=ωv(0) = \omega cm/s, the amplitude A=x(0)2+(v(0)ω)2=12+12=2A = \sqrt{x(0)^2 + \left(\frac{v(0)}{\omega}\right)^2} = \sqrt{1^2 + 1^2} = \sqrt{2} cm.

Chapter Concept:

Force Law in SHM

A.

3.1 Hz

B.

5.5 Hz

C.

8.7 Hz

D.

9.8 Hz
Correct Answer: C

Solution:

The frequency ff is given by f=12πkmf = \frac{1}{2\pi} \sqrt{\frac{k}{m}}. Assuming the mass m=3 kgm = 3 \text{ kg}, f=12π12003≈8.7 Hzf = \frac{1}{2\pi} \sqrt{\frac{1200}{3}} \approx 8.7 \text{ Hz}.

Chapter Concept:

Period and Frequency

A.

F=−kxF = -kx

B.

F=kxF = kx

C.

F=mω2xF = m\omega^2x

D.

F=−mω2xF = -m\omega^2x
Correct Answer: A

Solution:

In simple harmonic motion, the restoring force is proportional to the displacement from the equilibrium position and is directed towards the equilibrium position. Therefore, the correct expression is F=−kxF = -kx, where kk is the spring constant.

Chapter Concept:

Displacement, Amplitude, Phase and Angular Frequency

A.

10 cm/s

B.

20 cm/s

C.

31.4 cm/s

D.

62.8 cm/s
Correct Answer: C

Solution:

The maximum speed vmaxv_{\text{max}} in SHM is given by vmax=Aωv_{\text{max}} = A \omega. Substituting A=5 cmA = 5 \text{ cm} and ω=2π s−1\omega = 2 \pi \text{ s}^{-1}, we get vmax=5×2π=31.4 cm/sv_{\text{max}} = 5 \times 2 \pi = 31.4 \text{ cm/s}.

Chapter Concept:

Displacement, Amplitude, Phase and Angular Frequency

A.

ff

B.

2f2f

C.

4f4f

D.

2f\sqrt{2}f
Correct Answer: C

Solution:

The frequency of a mass-spring system is given by f=12πkmf = \frac{1}{2\pi} \sqrt{\frac{k}{m}}. If the spring constant kk is quadrupled and the mass mm is halved, the new frequency f′=12π4km/2=4ff' = \frac{1}{2\pi} \sqrt{\frac{4k}{m/2}} = 4f.

Chapter Concept:

Projection of Circular Motion

A.

Positive

B.

Negative

C.

Zero

D.

Cannot be determined
Correct Answer: B

Solution:

At point A, the particle is at one extreme of its motion, and the acceleration is directed towards the center of motion, hence negative.

Chapter Concept:

Periodic and Oscillatory Motions

A.

2πmk2\pi \sqrt{\frac{m}{k}}

B.

2πm2k2\pi \sqrt{\frac{m}{2k}}

C.

2π2mk2\pi \sqrt{\frac{2m}{k}}

D.

2πm4k2\pi \sqrt{\frac{m}{4k}}
Correct Answer: B

Solution:

When two identical springs are attached in parallel, the effective spring constant is 2k2k. Thus, the period of oscillation is T=2πm2kT = 2\pi \sqrt{\frac{m}{2k}}.

Chapter Concept:

Identifying SHM from Functions and Graphs

A.

ϕ=cos⁡−1(x0A)\phi = \cos^{-1}\left(\frac{x_0}{A}\right)

B.

ϕ=tan⁡−1(v0ωx0)\phi = \tan^{-1}\left(\frac{v_0}{\omega x_0}\right)

C.

ϕ=sin⁡−1(v0ωA)\phi = \sin^{-1}\left(\frac{v_0}{\omega A}\right)

D.

ϕ=tan⁡−1(ωx0v0)\phi = \tan^{-1}\left(\frac{\omega x_0}{v_0}\right)
Correct Answer: A

Solution:

The initial position x0=Acos⁡(ϕ)x_0 = A \cos(\phi) allows us to solve for the phase constant: ϕ=cos⁡−1(x0A)\phi = \cos^{-1}\left(\frac{x_0}{A}\right).

Chapter Concept:

Displacement, Amplitude, Phase and Angular Frequency

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True or False

Correct Answer: False

Solution:

The motion of a simple pendulum is simple harmonic only for small angular displacements. For larger angles, the approximation that sin⁡θ≈θ\sin \theta \approx \theta is not valid, and the motion deviates from simple harmonic.

Chapter Concept :

Simple Pendulum

Correct Answer: True

Solution:

The force law for SHM is given by F(t)=−kx(t)F(t) = -kx(t), where kk is the force constant, indicating that the force is proportional to the displacement and directed towards the equilibrium position.

Chapter Concept :

Force Law in SHM

Correct Answer: True

Solution:

The force law for simple harmonic motion is given by F(t)=−kx(t)F(t) = -kx(t), where kk is the force constant. This indicates that the force is proportional to the displacement x(t)x(t) and directed towards the equilibrium position.

Chapter Concept :

Force Law in SHM

Correct Answer: True

Solution:

The total mechanical energy in SHM remains constant and is calculated as E=12kA2E = \frac{1}{2} kA^2, where kk is the spring constant and AA is the amplitude of the motion.

Chapter Concept :

Energy in SHM

Correct Answer: True

Solution:

The period TT is indeed the time for one complete oscillation and is inversely related to the frequency vv, as given by the formula T=1vT = \frac{1}{v}.

Chapter Concept :

Period and Frequency

Correct Answer: True

Solution:

The formula x(t)=Acos⁡(ωt+Φ)x(t) = A \cos(\omega t + \Phi) accurately represents the displacement in SHM, with AA as the amplitude, ω\omega as the angular frequency, and Φ\Phi as the phase constant.

Chapter Concept :

Displacement, Amplitude, Phase and Angular Frequency

Correct Answer: False

Solution:

In simple harmonic motion, the force is always directed towards the center of motion, acting as a restoring force proportional to the displacement.

Chapter Concept :

Projection of Circular Motion

Correct Answer: True

Solution:

The velocity and acceleration expressions for SHM are periodic functions of time, as defined by the equations V(t)=−wAsin⁡(wt+Φ)V(t) = -wA \sin(wt + \Phi) and a(t)=−w2Acos⁡(wt+Φ)a(t) = -w^2A \cos(wt + \Phi), where ww is the angular frequency, AA is the amplitude, and Φ\Phi is the phase constant.

Chapter Concept :

Velocity and Acceleration in SHM

Correct Answer: False

Solution:

In simple harmonic motion, the kinetic energy is zero at the extreme positions of displacement and maximum at the mean position.

Chapter Concept :

Energy in SHM

Correct Answer: True

Solution:

The force in simple harmonic motion is described by F=−kxF = -kx, where kk is the force constant and xx is the displacement. This shows that the force is proportional to the displacement and directed towards the mean position.

Chapter Concept :

Spring-Mass System and Linear Oscillator

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