Tag: free, forced and damped oscillations

Questions Related to free, forced and damped oscillations

Frequency of oscillation of a body is $6\;Hz$ when force $F _1$ is applied and $8\;Hz$ when $F _2$ is applied. If both forces $F _1\;&\;F _2$ are applied together then, the frequency of oscillation is :

  1. $14\;Hz$

  2. $2\;Hz$

  3. $10\;Hz$

  4. $10\surd{2}\;Hz$


Correct Option: C
Explanation:

According to question,

$F _1=-K _1\,x\;\&\;F _2=-K _2\,x$

So $n _1=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{K _1}{m}}=6\;Hz;$

$n _2=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{K _2}{m}}=8\;Hz$

Now $F=F _1+F _2=-(K _1+K _2)x$

Therefore $n=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{K _1+K _2}{m}}$

$\Rightarrow n=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{4\pi^2\,n^2 _1\,m+4\pi^2\,n^2 _2\,m}{m}}$$=\sqrt{n _1^2+n _2^2}=\sqrt{8^2+6^2}$

$=10\;Hz$

The angular frequency of the damped oscillator is given by $\omega =\sqrt { \left( \dfrac { k }{ m } -\dfrac { { r }^{ 2 } }{ 4{ m }^{ 2 } }  \right)  }$ , where k is the spring constant, $m$ is the mass of the oscillator and $r$ is the damping constant. If the ratio $\dfrac { { r }^{ 2 } }{ mk }$ is $80$%, the change in time period compared to the undamped oscillator is approximately as follows:

  1. Decreases by $1$%

  2. Increases by $8$%

  3. Increases by $1$%

  4. Decreases by $8$%


Correct Option: A

In case of a forced vibration, the resonance wave becomes very sharp when the :

  1. Damping force is small

  2. Restoring force is small

  3. Applied periodic force is small

  4. Quality factor is small


Correct Option: A
Explanation:

In forced vibration, the resonance wave becomes very sharp when damping force is small (i.e. negligible)

The potential energy of a particle of mass 1 kg in motion along the x-axis is given by U = 4(1 - cos2x) J. Here x is in meter. The period of small oscillations (in sec) is _______.

  1. $2\pi$

  2. $\pi$

  3. $\dfrac{\pi}{2}$

  4. $\sqrt{2\pi}$


Correct Option: C
Explanation:
$M=1\ kg$
$U=u(1-\cos^2 x)$
OR $U=8\sin^2 x\quad [\because \ 1-\cos \theta =2\sin^2 \theta]$
$f=-\dfrac {du}{dx}$
$f=-u [0-(-\sin 2x) (2)]$
$f=-8\sin 2x\ N$
$a=-8\sin 2x \ m/ \sec^2 \quad [\because \ m=1\ kg]$
$\Rightarrow \ a\simeq -8(2x)$
$a=-16x$ [For small $x, \sin x \approx x$]
$\Rightarrow \ w^2 =4^2$
$w=4\ $ rad /ac
$T=2\dfrac {\pi}{w}=\dfrac {\pi}{2}\sec$

The period of oscillation of a simple pendulum of constant length is independent of

  1. size of the bob

  2. shape of the bob

  3. mass of bob

  4. all of these


Correct Option: D
Explanation:

$T=2 \pi \sqrt{\dfrac{L}{g}}$               (where L= length of sting)
From above equation, Time period only depend on the length of the string and g.

Option d 

Assertion : In damped oscillations, the energy of the system is dissipated continuously.
Reason : For the small damping, the oscillations remain approximately periodic.

  1. If both assertion and reason are true and reason is the correct explanation of assertion.

  2. If both assertion and reason are true and reason is not the correct explanation of assertion.

  3. If assertion is true but reason is false.

  4. If both assertion and reason are false.


Correct Option: B
Explanation:

In damped oscillation like the motion of simple pendulum swinging in air, motion dies out eventually. This is because of the air drag and the friction at the support oppose the motion of the pendulum and dissipate its energy gradually.

Assertion : In forced oscillations, the steady state motion of the particle is simple harmonic.
Reason : Then the frequency of particle after the free oscillations die out, is the natural frequency of the particle.

  1. If both assertion and reason are true and reason is the correct explanation of assertion.

  2. If both assertion and reason are true and reason is not the correct explanation of assertion.

  3. If assertion is true but reason is false.

  4. If both assertion and reason are false.


Correct Option: C
Explanation:

In forced oscillations, the frequency of particle after free oscillations die out, is the frequency of the driving force, not the natural frequency of the particle.

A force F= -4x-8 is acting on a block where x is position of block in meter. The energy of oscillation is 32 J, the block oscillate between two points Position of extreme position is:

  1. 6

  2. 0

  3. 4

  4. 3


Correct Option: A
Explanation:

$F=-4x-8\F=-4(x+2)\ \Rightarrow F=-4[x-(-2)]\F=-kx\rightarrow$ Distance mean position. So, this will be SHM

with $x=-2$ as mean position 
as $K _SHM=4$
Energy of SHM$=\cfrac{1}{2}KA^2=32\ \Rightarrow A^2=\cfrac{2\times32}{K}=\cfrac{2\times32}{4}=16\Rightarrow A=4$
and mean position $=-2$
Extereme position $=(-2+4)\quad and (-2-4)\=2\quad and\quad -6$

A block of $0.5\ kg$ is placed on a horizontal platform. The system is making vertical oscillations about a fixed point with a frequency of $0.5\ Hz.$ Find the maximum amplitude of oscillation if the block is not to lose contact with the horizontal platform?

  1. $0.6542\ m$

  2. $0.9927\ m$

  3. $0.7428\ m$

  4. $0.852\ m$


Correct Option: D

A highly rigid cubical block A of small mass M and side L is fixed rigidly on another cubical block B of the same dimensions and of low modulus of rigidity $\eta $ such that the lower face of A completely covers the upper face of B.  The lower face of B is rigidly held on horizontal surface.  A small force is applied perpendicular to the side faces of A.  After the force is withdrawn, block A executes small oscillations the time period of which is given by 

  1. $2\pi \sqrt{M\eta L}$

  2. $2\pi \sqrt{\frac{M-\eta }{L}}$

  3. $2\pi \sqrt{\frac{M-L}{\eta }}$

  4. $2\pi \sqrt{\frac{M-N}{\eta L}}$


Correct Option: A