Tag: waves

Questions Related to waves

Phase difference between a compression and its successful rarefaction is $2 \pi $radians

  1. True

  2. False


Correct Option: B
Explanation:

Phase difference between a crest and its successive trough is $ \pi $radians

A travelling wave has a velocity of 400 m/s and has a wavelength of 0.5 m. What is the phase difference between two points in the wave that are 1.25 milli secs apart

  1. $2 \pi$

  2. $2 \pi/3$

  3. $2 \pi/5$

  4. $ \pi/6$


Correct Option: A
Explanation:

The frequency of the wave is $f = v/\lambda=400/0.5=800 Hz$

Time period = 1/800 = 1.25 ms

Thus, both the points are one time period apart. Hence their phase difference will be zero or $2 \pi$

The correct option is (a)

Two waves of frequencies 20 Hz and 30 Hz travels out from a common point. The phase difference between them after 0.6 sec is

  1. $12\pi $

  2. ${\pi \over 2}$

  3. $\pi $

  4. ${{3\pi } \over 4}$


Correct Option: A
Explanation:

Beats period$=\cfrac{1}{30-20}=0.1second\ \Delta Y=\cfrac{2\pi}{T}\Delta t\ \quad=\cfrac{2\pi}{0.1}\times0.6=2\pi\times6=12\pi\ \therefore\Delta Y=12\pi$

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.