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 physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

Two SHMs are represented by the equations 
$y1=10sin(3\Omega t+\frac{\Omega }{4})$ and
$y2=5[sin3\Omega t+\sqrt{3}cos 3\Omega t]$. their amplitudes and in the ratio

  1. 1:2

  2. 2:1

  3. 1:3

  4. 1:1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Compare the given equations with,

$y=a\sin \omega t$

Where a is amplitude of wave

Here, a1 = 10

 a2 = 5

So the ratio is

$\dfrac{{{a} _{1}}}{{{a} _{2}}}=\dfrac{10}{5}=\dfrac{2}{1}$

Or $2:1$

Multiple choice physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

A simple harmonic oscillator starts from extreme position and covers a displacement half of its amplitude in a time '$t$', the further time taken by it to reach mean position is

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

$x=A cos(\omega t)$
Taking $t=0$ at extreme position.
$x=\dfrac{A}{2}$ is reached in time t
$\omega t=\dfrac{\pi }{3} ;\omega =\dfrac{\pi }{3t}$
$x=0$ is reached at $\omega t _{1}=\dfrac{\pi }{2}$
$t _{1}=\dfrac{\pi }{2\omega }=\dfrac{\pi \times 3t}{2\times\pi }=\dfrac{3t}{2}$
$t _{1}-t=\dfrac{t}{2}$

Multiple choice physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

The particle executes SHM on a straight line. At two positions its velocity $u$ and $v$ while acceleration, $\alpha$ and $\beta$ respectively $[\beta > \alpha >0]$, the distance between the two positions will be:-

  1. $\dfrac{u^2+v^2}{\alpha+\beta}$
  2. $-\dfrac{u^2-v^2}{\alpha+\beta}$
  3. $\dfrac{u^2-v^2}{\alpha-\beta}$
  4. $\dfrac{u^2+v^2}{\beta-\alpha}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that velocity is related as,

$u=\omega \sqrt{A^2-X _1^2}$
$v=\omega\sqrt{A^2-X _2^2}$
Acceleration is related as,
$\alpha=-\omega^2X _1^2$
$\beta=-\omega^2X _2^2$
$\dfrac{u^2-v^2}{u^2}=-(X _1^2-X _2^2)$
$=-(X _1+X _2)(X _1-X _2)$
$=\dfrac{\alpha+\beta}{u^2}(X _1-X _2)$
$X _1-X _2=-\dfrac{u^2-v^2}{\alpha+\beta}$

Multiple choice physics simple harmonic motion representing shm with circular motion shm as projection of circular motion simple harmonic motion (shm) as a projection of uniform circular motion

The particle executes SHM on a straight line. At two positions its velocity $u$ and $v$ while acceleration, $\alpha$ and $\beta$ respectively $[\beta > \alpha >0]$, the distance between the two positions will be:-

  1. $\frac{u^2+v^2}{\alpha+\beta}$
  2. $-\frac{u^2-v^2}{\alpha+\beta}$
  3. $\frac{u^2-v^2}{\alpha-\beta}$
  4. $\frac{u^2+v^2}{\beta-\alpha}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$u=w\sqrt{A^2-X _1^2}$
$V=w\sqrt{A^2-X _2^2}$
$\alpha=-w^2 x _1^2$
$\beta=-w^2 x _2^2$
$\frac{u^2-v^2}{u^2}=-(x _1^2-x _2^2)$
$=-(x _1+x _2)(x _1-x _2)$
$=\frac{(\alpha+\beta)}{w^2}(x _1-x _2)$
$x _1-x _2=\frac{u^2-v^2}{\alpha+\beta}$
Multiple choice force in shm oscillations oscillation and waves physics

Which one of the following equations of motion represents simple harmonic motion?

  1. Acceleration =$-k _0x+k _1x^2$
  2. Acceleration =$-k(x+a)$
  3. Acceleration =$k(x+a)$
  4. Acceleration =$kx$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Simple harmonic motion (SHM) is defined by a restoring force (or acceleration) that is directly proportional to the negative displacement from the equilibrium position, expressed as a = -kx. Option B represents this form, where the equilibrium position is shifted by 'a'.

Multiple choice force in shm oscillations oscillation and waves physics

The mass of particle executing S.H.M is 1 gm.If its periodic time is $\pi $ seconds, the value of force constant is:-

  1. 4 dynes/cm

  2. 4 N/cm

  3. 4 N/m

  4. 4 dynes/m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, mass $m=1 gm$; time period $T=\pi$ sec

The angular frequency $\omega=\frac{2\pi}{T}=\frac{2\pi}{\pi}=2$
As $\omega=\sqrt{k/m}$  where ($k$ is the force constant),
$k=m\omega^2=(1 gm)(2)^2=4$ dynes/cm

Multiple choice force in shm oscillations oscillation and waves physics

In SHM, select the wrong statement, where ${F}$ is the force, ${a}$ is the acceleration and ${v}$ is the velocity of the particle in SHM.

  1. $\overset{\rightarrow}{F}\times \overset{\rightarrow}{v}$ is a null vector
  2. $|\overset{\rightarrow}{F}\times \overset{\rightarrow}{a}|=0$ (always)
  3. $\overset{\rightarrow}{F}\times \overset{\rightarrow}{a}< {0}$
  4. $\overset{\rightarrow}{a}.\overset{\rightarrow}{x}<0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The velocity, acceleration and thus the force are along the direction of motion in the SHM.

So the cross product of Force with velocity or acceleration is always zero.
Thus $A$ and $B$ are correct.
Also, acceleration is always in opposite direction to the displacement vector. So dot-product of acceleration and displacement is always negative. Thus $D$ is also correct.
$C$ is the wrong statement as the cross product of force and acceleration is strictly zero, because, $F=ma\Rightarrow ma\times a=0$.

Multiple choice force in shm oscillations oscillation and waves physics

A 1 kg mass executes SHM with an amplitude 10 cm, it takes $2\pi$ seconds to go from one end to the other end. The magnitude of the force acting on it at any end is :

  1. 0.1 N

  2. 0.2 N

  3. 0.5 N

  4. 0.05 N
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As $  w  = \cfrac{2\pi}{T} = 1 \ rad/sec $ ;    Amplitude  $A  = 0.1 m$
magnitude of  force $ = m \times w^{2}.A$
                                  $=  0.1 N$

Multiple choice physics magnetic effects of electric current direct current types of current movement of charge

The function ${\sin}^{2}{\omega t}$ represents:

  1. a periodic, but not simple harmonic, motion with a period $2\pi/\omega$
  2. a periodic, but not simple harmonic, motion with a period $\pi/\omega$
  3. a simple harmonic motion with a period $2\pi /\omega$
  4. a simple harmonic motion with a period $\pi /\omega$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the identity sin^2(wt) = (1 - cos(2wt)) / 2, the motion is simple harmonic with an angular frequency of 2w, meaning the period is 2pi / (2w) = pi / w.

Multiple choice resonance oscillations physics

Three simple harmonic motions in the same direction having the same amplitude a and same period are superposed. If each differs in phase from the next by $45^o$, then.

  1. The resultant amplitude $(1+\sqrt{2})a$
  2. The phase of the resultant motion relative to the first is $90^o$
  3. The energy associated with the resulting motion is $(3+2\sqrt{2})$ times the energy associated with any single motion

  4. The resulting motion is not simple harmonic
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

Let $y _1=a\sin \left(\omega t-\cfrac {\pi}{4}\right)$
$y _2=a\sin (\omega t)$
$y _3=a\sin \left(\omega t+\cfrac {\pi}{4}\right)$
On super imposing, resulting SHM-
$y=a\left[\sin \left(\omega t-\cfrac{\pi}{4}\right)+\sin \omega t+\sin \left (\omega t+\cfrac {\pi}{4} \right)\right]$
$\implies y=a \left[2\sin \omega t\cos \cfrac {\pi}{4}+\sin \omega t\right]$
$\implies y=a(1+\sqrt {2})\sin \omega t$
$\therefore$ Resultant amplitude $=(1+\sqrt{2})a$
Also, $\cfrac {E _{resultant}}{E _{single}}=\left(\cfrac {A}{a}\right)^2$
$\implies \cfrac {E _{resultant}}{E _{single}}=(\sqrt{2}+1)^2$
$\implies \cfrac {E _{resultant}}{E _{single}}=(3+2\sqrt{2})$
$\therefore E _{resultant}=(3+2\sqrt{2})E _{single}$

Multiple choice resonance oscillations physics

The amplitude of damped oscillator becomes $\dfrac{1}{3}$ in $2\ s$. Its amplitude after $6\ s$ is $1/n$ times the original. The value of $n$ is ?

  1. $2^{3}$
  2. $3^{2}$
  3. $3^{1/2}$
  4. $3^{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The amplitude of a damped oscillator follows A(t) = A0 * exp(-bt). Given A(2) = A0 / 3, we have exp(-2b) = 1/3. For t = 6, A(6) = A0 * exp(-6b) = A0 * (exp(-2b))^3 = A0 * (1/3)^3 = A0 / 27. Thus, n = 27 = 3^3.

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

In forced oscillation of a particle the amplitude is maximum for a frequency $\omega _1$ of force, while the energy is maximum for a frequency $\omega _2$ of the force, then:

  1. $\omega _1= \omega _2$
  2. $\omega _1> \omega _2$
  3. $\omega _1 < \omega _2$ when damping is small and $\omega _1> \omega _2$ when damping is large
  4. $\omega _1< \omega _2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the amplitude of oscillation and energy to be maximum, the frequency of force must be equal to the initial frequency and this is only possible in resonance. In resonance state $ \omega _1 = \omega _2$.

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

The amplitude of a damped oscillator decreases to 0.9 times its original magnitude is 5 s .In another 10 s 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

For a damped oscillator, amplitude A(t) = A0 * e^(-bt). Given A(5) = 0.9 * A0, so e^(-5b) = 0.9. In another 10s (total 15s), A(15) = A0 * e^(-15b) = A0 * (e^(-5b))^3 = A0 * (0.9)^3 = 0.729 * A0.

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

A particle moves such that its acceleration is given by : $\alpha=-\beta(x-2)$
Here :$\beta$ is a positive constant and x the position from oigin. Time period of oscillations is:

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

The acceleration is given by a = -beta(x-2). This is the standard form of simple harmonic motion a = -omega^2(x-x0), where omega^2 = beta. The time period T is 2*pi/omega, which simplifies to 2*pi/sqrt(beta).

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

The time taken by a particle performing S.H.M. to pass from point $ A  $ to $  B  $ where its velocities are same is $2$ seconds. After another 2 seconds it returns to $ \mathrm{B}  $ . The time period of oscillation is (in seconds):

  1. $2$
  2. $4$
  3. $6$
  4. $8$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

According to the question, a and b points are such that they are

 located at same distances from the equilibrium position. 
Here also it is said that velocity is same 
,i.e; not only the magnitude but also the directions are same.
 So, total time period of oscillation $= 2×$( time taken to go from a to b
 + the next time taken to return at b) $= 2×(2+2)= 8$ sec.