Physics

Oscillations and Simple Harmonic Motion

138 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 power work and power work, energy and power physics energy and its forms

A time varying power $P=2t$ is applied on a particle of mass $m$. Find average power over a time interval from t=0 to t=t :

  1. $\displaystyle P _{av}= t$
  2. $\displaystyle P _{av}= 2t$
  3. $\displaystyle P _{av}= 4t$
  4. $\displaystyle P _{av}= 8t$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Average power over a time interval $t=0$ to $t=t$ is 
$\dfrac{\int _0^tPdt}{\int _0^tdt}$
$=\dfrac{\int _0^t 2tdt}{t}$
$=\dfrac{t^2}{t}=t$
Multiple choice physics oscillatory motion a few applications of linear shm simple pendulum example of simple harmonic motion

In a simple harmonic motion

  1. the potential energy is always equal to the kinetic energy

  2. the potential energy is never equal to the kinetic energy

  3. the average potential energy in any time interval is equal to the average kinetic energy in that time interval

  4. the average potential energy in one time period is equal to the average kinetic energy in this period.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In simple harmonic motion, the average kinetic energy and the average potential energy over one complete period are equal, both being half of the total energy.

Multiple choice physics oscillations a few applications of linear shm simple pendulum example of simple harmonic motion

A particle executes $SHM$ with a time period of $16\ s$. At time $t=2\ s$, the particle crosses the mean position while at $t=4s$, its velocity is $4ms^{-1}$. The amplitude of motion in meter is:

  1. $\sqrt{2}\pi$
  2. $16\sqrt{2} \pi$
  3. $ \dfrac{32\sqrt{2}}{\pi}$
  4. $ \dfrac{4}{\pi}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the equation of $S.H.M$ is:-


$x=a\sin\left(\dfrac{2\pi }{T}t+\phi\right)$

when $t=2s, x=0$ and $T=16s$ So,

$0=a\sin \left(\dfrac{\pi}{4}+\phi\right)$

Or $\phi=-\dfrac{\pi}{4}$

Therefore the eqn of $S.H.M$ is:-

$x=a\sin =\left(\dfrac{2\pi}{T}t-\dfrac{\pi}{4}\right)$

Now at time $t=4s, V=4m/s$

 So
$V=d\times dt=a\times \dfrac{2\pi}{T}\cos\left(\dfrac{2\pi}{T}t-\dfrac{\pi}{4}\right)$

So, $4=a\times \dfrac{2\pi}{16}\cos\left(\dfrac{\pi}{2}-\dfrac{\pi}{4}\right)$

Or, $4=a\times \dfrac{\pi}{8}\times \dfrac{1}{\sqrt{2}}$

Or $a=\dfrac{32\sqrt{2}}{\pi}$

Hence option $C$ is correct

Multiple choice physics oscillations a few applications of linear shm simple pendulum example of simple harmonic motion

The particle is executing S.H.M. on a line 4 cms long. If its velocity at its mean position is 12 cm/sec, its frequency in Hertz will be :

  1. $\dfrac{2\pi}{3}$
  2. $\dfrac{3}{2\pi}$
  3. $\dfrac{\pi}{3}$
  4. $\dfrac{3}{\pi}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given,


$A=4cm$


$v=12cm/s$ at $x=0$ mean position

The velocity of particle performing S.H.M is given by

$v=\omega \sqrt{A^2-x^2}$

$12=\omega \sqrt{4^2-0}$

$12=4\omega$

$\omega =2\pi f=3$

$f=\dfrac{3}{2\pi}$

The correct option is B.

Multiple choice physics oscillatory motion a few applications of linear shm simple pendulum example of simple harmonic motion

A function of time given by $\left(\sin{\omega t}-\cos{\omega t}\right)$ represents

  1. simple harmonic motion

  2. non-periodic motion

  3. periodic but not simple harmonic motion

  4. oscillatory but not simple harmonic motion
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} \sin  \omega t-\cos  \omega t \ =\sqrt { 2 } \left[ { \frac { 1 }{ { \sqrt { 2 }  } } \sin  \omega t-\frac { 1 }{ { \sqrt { 2 }  } } \cos  \omega t } \right]  \ =\sqrt { 2 } \left[ { \sin  \omega t\times \cos  \frac { \pi  }{ 4 } -\cos  \omega t\times \sin  \frac { \pi  }{ 4 }  } \right]  \ =\sqrt { 2 } \sin  \left( { \omega t-\frac { \pi  }{ 4 }  } \right)  \ this\, \, function\, \, represents\, \, SHM\, \, as\, \, it\, \, can\, \, be\, \, written\, \, in\, \, the\, \, form: \ a\sin  \left( { \omega t+\phi  } \right)  \ its\, \, period\, \, is,\, \, \frac { { 2\pi  } }{ \omega  }  \end{array}$

Hence,
option $(A)$ is correct answer.

Multiple choice physics oscillatory motion a few applications of linear shm simple pendulum example of simple harmonic motion

A particle is subjected to two simple harmonic motions along $x$ and $y$ directions according to $x=3\sin\ 100\pi t$ $y=4\sin\ 100\pi t$

  1. Motion of particle will be on ellipse travelling in clockwise direction.
  2. Motion of particle will be on a straight line with slope $4/3$
  3. Motion will be simple harmonic motion with amplitude $5$.
  4. Phase difference between two motions is $\pi/2$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics oscillatory motion a few applications of linear shm simple pendulum example of simple harmonic motion

The equation of motion of a particle of mass $1$ g is $\frac{{{d^2}x}}{{d{t^2}}} + {\pi ^2}x = 0$ where $x$ is displacement (in m) from mean position. The frequency of oscillation is ( in Hz):

  1. $\frac{1}{2}$

  2. 2
  3. $5\sqrt {10} $
  4. $\frac{1}{{5\sqrt {10} }}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation is d^2x/dt^2 + pi^2 * x = 0. Comparing to d^2x/dt^2 + w^2 * x = 0, we get w^2 = pi^2, so w = pi. Since w = 2 * pi * f, then pi = 2 * pi * f, which gives f = 1/2 Hz.

Multiple choice physics simple harmonic motion a few applications of linear shm simple pendulum example of simple harmonic motion

Suppose a tunnel is dug along a diameter of the earth. A particle is dropped from a point, a distance $h$ directly above the tunnel, the motion of the particle is

  1. Simple harmonic

  2. Parabolic

  3. Oscillatory

  4. Periodic
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

When a particle is dropped from a height $h$ above the centre of tunnel.
$(i)$ It will oscillate, through the earth to a height $h$ on both sides
$(ii)$ The motion of particle is periodic
$(iii)$ The motion of particle will not be $SHM$.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

The phase difference between the particle at one compression and another particle in third compression is

  1. $\pi $ radians
  2. $2\pi $ radians
  3. $3\pi $ radians
  4. $4\pi $ radians
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

phase difference between two successive compression is $2\pi $
$\therefore $ phase difference between a particle at one compression and in third compression is $2(2\pi)= 4\pi $

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

Two $SHMs$ are given by $Y _{1}= a\left[ \sin { \left( \dfrac { \pi  }{ 2 }  \right)  } t+\varphi  \right]$ and $Y _{2}= b\sin { \left[ \left( \dfrac { 2\pi t }{ 3 }  \right) +\varphi  \right]  }$ . The phase difference between these two after $'1'\ sec$ is:

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

At t=1, the phase of Y1 is pi/2 + phi. The phase of Y2 is 2pi/3 + phi. The difference is 2pi/3 - pi/2 = 4pi/6 - 3pi/6 = pi/6.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

Two particles are executing simple harmonic motion of the same amplitude $A$ and frequency $\omega$ along the $x-axis.$ Their mean position is separated by distance $X _0(X _0 > A)$. If the maximum separation between them is $(X _0 + A ),$ the phase difference between their motion is :-

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

${X _1} = A\sin \left( {wt + {\phi _1}} \right)$

${X _2} = A\sin \left( {wt + {\phi _2}} \right)$
${X _1} - {X _2} = A\left[ {2\sin \left[ {wt + \frac{{{\phi _1}}}{{{\phi _2}}}} \right]\sin \left[ {\frac{{{\phi _1} - {\phi _2}}}{2}} \right]} \right]$
$A = 2A\sin \left( {\frac{{{\phi _1} - {\phi _2}}}{2}} \right)$
$\frac{{{\phi _1} - {\phi _2}}}{2} = \frac{\pi }{6}$
${\phi _1} = \frac{\pi }{3}$
Hence,
option $(D)$ is correct answer.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

Two particles executing SHM of same frequency, meet at x=+A/2, while moving in opposite directions. Phase difference between the particles is 

  1. $\frac{\pi}{6}$
  2. $\frac{\pi}{3}$
  3. $\frac{5\pi}{6}$
  4. $\frac{2\pi}{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

At x = A/2, the phase angles are pi/6 and 5pi/6. Since they move in opposite directions, one is at pi/6 (moving away from equilibrium) and the other is at 5pi/6 (moving toward equilibrium). The difference is 5pi/6 - pi/6 = 4pi/6 = 2pi/3.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

Equation ${ y } _{ 1 }=0.1sin\left( 100\pi t+\dfrac { \pi  }{ 3 }  \right) $ and ${ y } _{ 2 }=0.1$ cos $\pi t$ The phase difference of the velocity of particle 1, with respect to the velocity of particle 2 is 

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

v1 = dy1/dt = 0.1 * 100pi * cos(100pi*t + pi/3). v2 = dy2/dt = -0.1 * pi * sin(pi*t). This question is garbled regarding the frequencies (100pi vs pi), making a standard phase difference calculation between them invalid.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

Two simple harmonic motions are represented by the equations 
$y _1=10\sin \left(3\pi t+\dfrac{\pi}{4}\right)$
and $y _2=5(3\sin 3\pi t+\sqrt 3 \cos 3\pi t)$ Their amplitudes are in the ratio of :

  1. $\sqrt 3$
  2. $1/\sqrt 3$
  3. $2$
  4. $1/6$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

y1 = 10 sin(3pi*t + pi/4), amplitude = 10. y2 = 5(3 sin(3pi*t) + sqrt(3) cos(3pi*t)). Using R = sqrt(A^2 + B^2 + 2AB cos(phi)), y2 = 5 * sqrt(3^2 + sqrt(3)^2) * sin(...) = 5 * sqrt(9+3) = 5 * sqrt(12) = 10 * sqrt(3). Ratio = 10 / (10 * sqrt(3)) = 1/sqrt(3).