Tag: oscillation and waves

Questions Related to oscillation and waves

A progressive wave of wavelength 5 cm moves along +X axis. What is the phase difference between two points on the wave separated by a distance of 3 cm at any instant

  1. $3 \pi/5$

  2. $6 \pi/5$

  3. $2 \pi/5$

  4. $7 \pi/5$


Correct Option: B
Explanation:

Phase difference = $(2\pi/5)$ path difference $\implies$ phase difference = $6 \pi/5$

The correct option is (b) 

The phase difference between two waves, represented by ${ y } _{ 1 }={ 10 }^{ -6 }\sin { \left[ 100t+\left( x/50 \right) +0.5 \right]\ m } $ and ${ y } _{ 2 }={ 10 }^{ -6 }\cos { \left[ 100t+\left( x/50 \right)  \right]\ m } $. Where $x$ is expressed in metre and $t$ is expressed in seconds, is approximately

  1. $1.07\ radian$

  2. $2.07\ radian$

  3. $0.5\ radian$

  4. $1.5\ radian$


Correct Option: A

The irreducible phase difference in any wave of 5000 A from a source of light is

  1. $\pi$

  2. $12\pi$

  3. $12\pi \times{10}^{6}$

  4. $\pi \times {10}^{6}$


Correct Option: A

In a string the speed of wave is $10m/s$ and its frequency is $100$ Hz. The value of the phase difference at a distance $2.5$cm will be :

  1. $\pi/2$

  2. $\pi/8$

  3. $3\pi/2$

  4. $4\pi$


Correct Option: A
Explanation:

Given,


$f=100Hz$


$x=2.5cm$

$v=10m/s$  path difference

wavelength, $\lambda =\dfrac{v}{f}=\dfrac{10}{100}=0.1m$

The phase difference, $\phi=\dfrac{2\pi}{\lambda}\times x$

$\phi=\dfrac{2\pi}{0.1}\times 2.5\times 10^{-2}$

$\phi=\dfrac{\pi}{2}$

The correct option is A.

A transverse progressive wave on a stretched string has a velocity of $10ms^{-1}$ and frequency of $100Hz$. The phase difference between two particles of the string which nbare $2.5cm$ apart will be :

  1. $\cfrac{\pi}{8}$

  2. $\cfrac{\pi}{4}$

  3. $\cfrac{3\pi}{8}$

  4. $\cfrac{\pi}{2}$


Correct Option: D
Explanation:

Given,


$v=10m/s$ velocity


$f=100Hz$ frequency

$\Delta=2.5cm$ path difference

wavelength, $\lambda=\dfrac{v}{f}$

$\lambda=\dfrac{10}{100}=0.1m$

Phase difference,

$\phi=\dfrac{2\pi}{\lambda}\times \Delta$

$\phi=\dfrac{2\pi}{0.1}\times 2.5\times 10^{-2}$

$\phi=\dfrac{\pi}{2}$

The correct option is D.

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. $\dfrac {\pi}{6}$


Correct Option: D

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


Correct Option: D
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.

The distance between two consecutive crests in a wave train produced in string is 5 m. If two complete waves pass through any point per second, the velocity of wave is:

  1. $2.5 \mathrm { m } / \mathrm { s }$

  2. $5 \mathrm { m } / \mathrm { s }$

  3. $10 \mathrm { m } / \mathrm { s }$

  4. $15 \mathrm { m } / \mathrm { s }$


Correct Option: C
Explanation:

Given,

Two consecutive crest =$5\,m$.

Two wave cross in time, $t=\,1\sec $

Wavelength = distance between two consecutive crest

$\lambda =5\,m$

Velocity, $v=\dfrac{2\lambda }{t}=\dfrac{2\times 5}{1}=10\,m{{s}^{-1}}$

Hence, wave speed is $10\,m{{s}^{-1}}$

 

Two waves $E _ { 1 } = E _ { 0 } \sin \omega t$  and $E _ { 2 } = E _ { 0 } \sin ( \omega t + 60 )$ superimpose each other. Find out initial phase of resultant wave?

  1. $30 ^ { \circ }$

  2. $60 ^ { \circ }$

  3. $120 ^ { \circ }$

  4. $0 ^ { \circ }$


Correct Option: A
Explanation:
Let the resultant wave be 
$E=E0' \sin (wt+\phi)$./ Then 
$E=E _1+E _2$
$E=E _0\sin wt +E _0\sin (wt+60^o)$
$E=E _0(\sin wt +\sin (wt +60^o)$
As $\sin \cos +\sin (B)=2\sin \left (\dfrac {A+B}{2}\right) \cos \left (\dfrac {B-A}{2}\right) $
$\sin wt +\sin (wt +60^o)=2 \sin (wt +30^o).\cos (wt)$
So, $E=2E _0 \cos (wt) \sin (wt+30^o)$
So, $E _0=2E _0 \cos wt $ and $ \phi =30^o$
Option $A$ is correct


If the frequency of ac is 60 Hz the time difference corresponding to a phase difference of ${ 60 }^{ \circ  }$ is 

  1. 60 s

  2. 1 s

  3. $\dfrac { 1 }{ 60 } s$

  4. $\dfrac { 1 }{ 360 } s$


Correct Option: D