Tag: physics

Questions Related to physics

A particle of mass m is in one dimensional potential field and its potential energy is given by the following equation U(x)=${U _0}\left( {1 - \cos \;aX} \right)\;where\;{U _0}$ and $\alpha $ constants.The period of the particle for small oscillations near the equilibrium will be-

  1. $2\pi \sqrt {\dfrac{{m{\alpha ^2}}}{{m{\alpha ^2}{U _0}}}} $

  2. $2\pi \sqrt {m{\alpha ^2}{U _0}} $

  3. $2\pi \sqrt {\dfrac{m}{{{\alpha ^2}{U _0}}}} $

  4. $2\pi \sqrt {\dfrac{{{\alpha ^2}{U _0}}}{m}} $


Correct Option: A

A boy is playing on a swing in sitting position. the time period of oscillation of the swing is T, if the boy stands up, the time period of oscillation of the spring will be:

  1. $T$

  2. $ Less \ than T$

  3. $More\  than T$

  4. such as cannot be predicted


Correct Option: A

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. $8$


Correct Option: D
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.

A student measures the time period of oscillation of a simple pendulum. He uses the data to estimate the acceleration due to gravity 9g) at that place. If the maximum percentage error in measurement of length pendulum and that in time are $ e _{1} $ and $ e _{2} $ respectively then percentage error estimation of ''g'' is :

  1. $

    e _{1}+2 e _{2}

    $

  2. $

    2 e 1+e 2

    $

  3. $

    e 1+e _{2}

    $

  4. $

    e 1-e _{2}

    $


Correct Option: A

The phase of particle in SHM is found to increase by $14 \pi$ in 3.5 sec. Its frequency of oscillation is

  1. $2 Hz$

  2. $1/2 Hz$

  3. $1 Hz$

  4. $2 \pi Hz$


Correct Option: B

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