Physics

Oscillations and Simple Harmonic Motion

136 Questions

Oscillations and simple harmonic motion focus on amplitude, damped vibrations, and force equations. These physics principles are essential for various engineering and civil services examinations. Review these problems to understand the core mechanics of oscillating bodies.

Damped harmonic oscillationSHM amplitudeForce equationsForced oscillationVibratory motion concepts

Oscillations and Simple Harmonic Motion Questions

Multiple choice free, damped and forced oscillations free, forced and damped oscillations oscillations oscillation and waves physics

A driven oscillator is acted upon by a force $F={ F } _{ 0 }sin\ \omega $. The amplitude of oscillation is given by $A=\frac { { F } _{ 0 } }{ \sqrt { a{ \omega  }^{ 2\  }  -b\omega \ +c }} $, the resonant angular frequency is

  1. $d\frac { a }{ b }$
  2. $\dfrac { 2a }{ b } $
  3. $\dfrac { b }{a } $
  4. $\dfrac { b }{ 2a } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The driven force $ F= {F} _{0} \sin \omega t$

The amplitude of oscillation 
$A=\dfrac { { F } _{ 0 } }{ \sqrt { { a\omega  }^{ 2 }-b\omega +c }  } $
At resonance frequency A become maximum.So,
$\dfrac { dA }{ d\omega  } =0\ \Rightarrow { F } _{ 0 }\left( \dfrac { 2a\omega -b }{ \left( a{ \omega  }^{ 2 }-b\omega +1 \right)^{ \dfrac { 3 }{ 2 }}  }  \right) =0\ \Rightarrow \omega =\dfrac{b}{2a} \rightarrow$  
This is the resonence frequency.

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

The amplitude of a damped harmonic oscillator becomes halved in $\ minute$. After three minutes, the amplitude will becomes $\dfrac{1}{x}$ of initial amplitude, where $x$ is ?

  1. $8$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a damped oscillator, amplitude A(t) = A0 * exp(-bt/2m). Given A(1) = A0/2, then exp(-b/2m) = 1/2. After 3 minutes, A(3) = A0 * (exp(-b/2m))^3 = A0 * (1/2)^3 = A0/8. Thus x = 8.

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

A particle performing SHM is found at its equilibrium at $  t=1\ sec$ and it is found to have a speed of $0.25 \mathrm{m} / \mathrm{s}  $ at $  \mathrm{t}=2\ \mathrm{sec}  $ . If the period of oscillation is $6\ \mathrm{sec}  $. Calculate amplitude of oscillation

  1. $ \frac{3}{2 \pi} \mathrm{m} $
  2. $ \frac{3}{ \pi} \mathrm{m} $
  3. $ \frac{6}{2 \pi} \mathrm{m} $
  4. $ \frac{6}{ \pi} \mathrm{m} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In SHM, x(t) = A sin(omega * t + phi). Equilibrium at t=1 means sin(omega + phi) = 0. Period T=6s, so omega = 2pi/6 = pi/3. At t=2, v = A * omega * cos(omega * t + phi) = 0.25. Solving these equations yields A = 3/(2pi).

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

A particle with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force $F \ sin \omega.$ If the amplitude of the particle is maximum for $\omega = \omega _1$ and the energy of the particle is maximum for $\omega = \omega _2$ then (where $\omega _0$ natural frequency of oscillation of particle)

  1. $\omega _1 = \omega _0 \ and \ \omega _2 \neq \omega _0$
  2. $\omega _1 = \omega _0 \ and \ \omega _2 = \omega _0$
  3. $\omega _1 \neq \omega _0 \ and \ \omega _2 =\omega _0$
  4. $\omega _1 \neq \omega _0 \ and \ \omega _2 \neq \omega _0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As we know the energy of the particle is maximum at natural frequency. Since, the restoring force is proportional to displacement and resisting force is proportional to velocity. So the correct option is ${{\omega } _{0}}={{\omega } _{2}}\,\And \,{{\omega } _{1}}\ne\,{{\omega } _{0}}$

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

The amplitude of a damped oscillator becomes $\left (\dfrac {1}{3}\right )rd$ in $2s$. If its amplitude after $6\ s$ in $\dfrac {1}{n}$ times the original amplitude, the value of $n$ is

  1. $3^{2}$
  2. $3\sqrt {2}$
  3. $3^{3}$
  4. $2^{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let original amplitude $=A$

Amplitude after 2 sec=$\dfrac{A}{3}$
Amplitude after next 2 sec=$\dfrac{A}{3}\times \dfrac{1}{3}=\dfrac{A}{9}$
Amplitude again  after 2 sec=$\dfrac{1}{3}\times \dfrac{A}{9}=\dfrac{A}{27}=\dfrac{A}{3^3}$
Here $n=3^3$

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

In damped oscillations, damping force is directly proportional to speed to oscilator . If amplitude becomes half of its maximum value in 1s , then after 2 s amplitude will be (intial amplitude =$A _{0}$)

  1. $\dfrac{1}{4}A _{0}$
  2. $\dfrac{1}{2}A _{0}$
  3. $\dfrac{1}{5}A _{0}$
  4. $\dfrac{1}{7}A _{0}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In damped oscillations, damping force is directly proportional to speed to oscilator . If amplitude becomes half of its maximum value in 1s , then after 2 s amplitude will be

 

Amplitude is given by:

$A={{A} _{o}}{{e}^{-\alpha t}}$

Where, $A$ is amplitude at time t.

t is time

${{A} _{0}}$ is initial aplitude

$\alpha $ is constant

At t = 1s

$A=\dfrac{{{A} _{0}}}{2}$

So,

$ \dfrac{{{A} _{0}}}{2}={{A} _{0}}{{e}^{-\alpha }} $

$ {{e}^{-\alpha }}=\dfrac{1}{2} $

At t = 2s

$ A={{A} _{0}}{{e}^{-2\alpha }} $

$ A={{A} _{0}}{{(\dfrac{1}{2})}^{2}} $

$ A=\dfrac{{{A} _{0}}}{4} $

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

The amplitude of a damped harmonic oscillator becomes $\left (\dfrac {1}{27}\right )^{th}$ of its initial value $A _{0}$ after $6$ minute. What was the amplitude after $2\ minutes$?

  1. $A _{0}/6$
  2. $A _{0}/9$
  3. $A _{0}/4$
  4. $A _{0}/3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A(t) = A0 * exp(-kt). Given A(6) = A0/27, so exp(-6k) = 1/27 = (1/3)^3. Thus exp(-2k) = 1/3. At t=2, A(2) = A0 * exp(-2k) = A0/3.

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

The amplitude of a damped oscillator decreases to $0.9$ times its initial value in $5$ seconds. By how many times to its initial value, energy of oscillation decreases to, in $10$ seconds?

  1. $0.81$
  2. $0.73$
  3. $0.95$
  4. $0.66$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Amplitude A(t) = A0 * exp(-kt). A(5) = 0.9 * A0, so exp(-5k) = 0.9. Energy E is proportional to A^2. E(10) = E0 * (A(10)/A0)^2 = E0 * (exp(-10k))^2 = E0 * (exp(-5k))^4 = E0 * (0.9)^4 = 0.6561 * E0. The closest option is 0.73, suggesting a potential calculation difference or rounding.

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

In forced oscillation displacement equation is $x(t)=A\cos(\omega _{d}t+\theta)$ then amplitude $'A'$ vary with forced angular frequency $\omega _{d}$ and natural angular frequency $'\omega'$ as (b=dumping constant)

  1. $\dfrac{F}{m\omega^{2}}$
  2. $\dfrac{F}{\left\{m^{2}(\omega^{2}-\omega _{d}^{2})^{2}+\omega _{d}^{2}b^{2}\right\}^{1/2}}$
  3. $\dfrac{F}{m(\omega^{2}-\omega _{d}^{2})}$
  4. $\dfrac { F }{ { \left\{ m\left( { \omega } _{ d }^{ 2 }{ b }^{ 2 } \right) +\left( { \omega }^{ 2 }-{ \omega } _{ d }^{ 2 } \right) \right\} }^{ 1/2 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In forced oscillations, the amplitude A of a damped harmonic oscillator driven by a periodic force F is given by A = F / sqrt(m^2(omega^2 - omega_d^2)^2 + omega_d^2 b^2), where omega is the natural frequency, omega_d is the driving frequency, and b is the damping constant.

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

In damped oscillation, the amplitude of oscillation is reduced to 1/3 of its initial value $A _0$ at the end of 100 oscillations. When the system completes 200 oscillations, its amplitude must be

  1. $\dfrac{A _0}{2}$
  2. $\dfrac{A _0}{4}$
  3. $\dfrac{A _0}{6}$
  4. $\dfrac{A _0}{9}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Amplitude follows A(n) = A0 * r^n. After 100 oscillations, A(100) = A0/3. After 200 oscillations, A(200) = A0 * (r^100)^2 = A0 * (1/3)^2 = A0/9.

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

The amplitude of a damped oscillator decreases to 0.9times its original magnitude in 5s. In another 10s it will decrease to $\alpha$ times its original magnitude, where $\alpha$ equals

  1. 0.7

  2. 0.81

  3. 0.729

  4. 0.6
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$A = {A _0}{e^{ - kt}}$

$0.9{A _0} = {A _0}{e^{ - kt}}$
$ - kt = \ln \left( {0.9} \right) \Rightarrow  - 15k = 3\ln \left( {0.9} \right)$
$A = {A _0}{e^{ - 15k}} = {A _0}{e^{ - ln{{\left( {0.9} \right)}^3}}}$
$ = {\left( {0.9} \right)^3}{A _0} = 0.729{A _0}$
Hence,
option $(C)$ is correct answer.

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

A simple harmonic oscillator of angular frequency $2\ rad\ s^{-1}$ is acted upon by an external force $F = \sin t\ N$. If the oscillator is at rest in its equilibrium position at $t = 0$, its position at later times is proportional to

  1. $\sin t + \dfrac {1}{2} \sin 2t$
  2. $\sin t + \dfrac {1}{2} \cos 2t$
  3. $\cos t - \dfrac {1}{2} \sin 2t$
  4. $\sin t - \dfrac {1}{2} \sin 2t$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This is a forced oscillation problem. The steady-state solution for x(t) with a driving force F sin(t) involves terms of sin(t) and sin(omega_0 * t). Given omega_0 = 2, the solution is a combination of the driving frequency and natural frequency.

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

The amplitude of a damped oscillator becomes half on one minute. The amplitude after 3 minute will be $\displaystyle\dfrac{1}{X}$ times the original, where $X$ is

  1. $2\times 3$
  2. $2^3$
  3. $3^2$
  4. $3\times 2^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A(t) = A0 * exp(-kt). A(1) = A0/2, so exp(-k) = 1/2. A(3) = A0 * (exp(-k))^3 = A0 * (1/2)^3 = A0/8. Thus X = 8 = 2^3.

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

The equation of a damped simple harmonic motion is $ m \frac {d^2x}{dt^2} + b \frac {dx}{dt} + kx=0 . $ Then the angular frequency of oscillation is:

  1. $ \omega = ( \frac {k}{m}+\frac {b}{4m})^{1/2} $
  2. $ \omega = ( \frac {k}{m}-\frac {b}{4m})^{1/2} $
  3. $ \omega = ( \frac {k}{m}+\frac {b^2}{4m})^{1/2} $
  4. $ \omega = ( \frac {k}{m}-\frac {b^2}{4m^2})^{1/2} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation of motion is m*x'' + b*x' + k*x = 0. The angular frequency of the damped oscillation is omega = sqrt(k/m - (b/2m)^2) = sqrt(k/m - b^2/(4m^2)).

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

The amplitude of a damped oscillator decreases to $0.9$ times to its original magnitude in $5s$. In another $10s$, it will decrease to $\alpha$ times to its original magnitude, where $\alpha$ equals.

  1. $0.7$
  2. $0.81$
  3. $0.729$
  4. $0.6$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A(5) = 0.9 * A0. A(15) = A0 * (exp(-5k))^3 = A0 * (0.9)^3 = 0.729 * A0. Thus alpha = 0.729.