Tag: wave motion

Questions Related to wave motion

A string is properly tuned:

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. When the beat frequency vanishes.

  2. When the beat frequency is maximum.

  3. When the beat frequency is minimum.

  4. When the beat frequency is between maximum and minimum.

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Correct Option: B

A heavy flexible rope hangs vertically. The speed of a transverse wave at a height $h$ from the free end is

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. $\sqrt { g h }$

  2. $\sqrt { g / h }$

  3. $\sqrt { 2 g h }$

  4. $\sqrt { h / g }$

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Correct Option: A

Small amplitude progressive waves in a stretched string have a speed of 100 cm/s and frequency 100 Hz. The phase difference between two points 2.75 cm apart on the string, in radians is 

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. $\dfrac { \pi }{ 4 } $

  2. $\dfrac { 3\pi }{ 4 } $

  3. $0$

  4. $\dfrac { 11\pi }{ 4 } $

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Correct Option: D

A tension in wire is 40N and 10 m of wire has a mass of 0.01 kg . The speed of transverse waves in m/s in the wire is :

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. 200

  2. 80

  3. 300

  4. 180

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Correct Option: A
Explanation:

We know, Speed of transverse wave $(v) = \sqrt{\dfrac{T}{\mu}}$


where, T = Tension = 40N and  $\mu = $ mass per unit length = $\dfrac{0.01}{10} = 10^{-3}\; kg/m$ 

$\Rightarrow v = \sqrt{\dfrac{40}{10^{-3}}} = 200 m/s$

Therefore, A is correct option.

A string of mass $2.5\ kg$ is under a tension of $200\ N$. The length of the stretched string is $20.0\ m$. If the transverse jerk is struck at one end of the string, the disturbance will reach the other end in

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. One second

  2. $0.5$ second

  3. $2\ seconds$

  4. Data given is insufficient

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Correct Option: B

The  equation of a transverse wave travel on a rope is given   y = 10 sin $\pi$(0.01x - 2.00t) where y and x in cm and t in seconds.The maximum transverse  speed  of a particle in the rope about 

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. 62.8 cm / s

  2. 75 cm / s

  3. 100 cm / s

  4. 121 cm / s

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Correct Option: A

A wave represented by equation $y = 2(mm) \, sin \, [4 \pi (sec^{-1}) t - 2 \pi (m^{-1}) X]$ is superimposed with another wave $y = 2 (mm) sin [4 \pi (sec^{-1}) t + 2 \pi (m^{-1}) x + \pi/3]$ on a tight string.
Phase difference between two particles with are located at $x _1 = 1/7$ and $x _2 = 5/12$ is :

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. $0$

  2. $\dfrac{5 \pi}{6}$

  3. $\pi$

  4. $\dfrac{5 \pi}{3}$

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Correct Option: C
Explanation:

${y _1} = 2\sin \left[ {4\pi t - 2\pi x} \right]$

${y _2} = 2\sin \left[ {4\pi t - 2\pi x} \right]$
$y = {y _1} + {y _2} = 2\left[ {2\sin 4\pi t\,\,\cos 4\pi x} \right]$
$ = 4\sin 4\pi t\cos 4\pi x$
$y = 4\cos 4\pi \sin 4\pi t$
Amp pass$\left| {4\cos 4\pi x} \right| = 4$
$ \Rightarrow \cos 4\pi x =  \pm 1$
$ = 0,\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{4}{4},\frac{5}{4},\frac{6}{4}.......$
$\therefore for\,{x _1} = \frac{1}{9}and\,{x _2} = \frac{5}{{12}}$
hence,
phase difference is$\pi$

A travelling wave on a string is given by $y = A$ $A \sin \left[ \alpha x + \beta t + \cfrac { \pi } { 6 } \right]$ The displacement and velocity of oscillation of a point $\alpha =$ $0.56 / \mathrm { cm } , \beta = 12 / \mathrm { sec }$ $A = 7.5 \mathrm { cm } , x = 1$ $\mathrm { cm }$ and $\mathrm { t } = 1 \mathrm { s }$ is

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. $4.6 \mathrm { cm } , 46.5 \mathrm { cm } s ^ { - 1 }$

  2. $3.75 \mathrm { cm } , 77.94 \mathrm { cm } \mathrm { s } ^ { - 1 }$

  3. $1.76 \mathrm { cm } , 7.5 \mathrm { cms } ^ { - 1 }$

  4. $7.5 \mathrm { cm } , 75 \mathrm { cm } \mathrm { s } ^ { - 1 }$

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Correct Option: B
Explanation:

$\begin{array}{l} y=7.5\sin  \left[ { 0.56\alpha +12t+\dfrac { \pi  }{ 6 }  } \right]  \ at\, \, x=1 \ & \, \, t=1 \ y=7.5\sin  \left[ { 0.56+12+\dfrac { \pi  }{ 6 }  } \right]  \ =7.5\sin  \left[ { 12.56+\dfrac { \pi  }{ 6 }  } \right]  \ =7.5\sin  \left[ { 4\pi +\dfrac { \pi  }{ 6 }  } \right]  \ =3.75\, \, cm \end{array}$

$\therefore$ Option $B$ is correct .

A sine wave is travelling in a medium. The minimum distance between the two particles. always having same speed is 

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. $\lambda / 4$

  2. $\lambda / 3$

  3. $\lambda / 2$

  4. $\lambda $

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Correct Option: C
Explanation:

In a sine wave particles that are separated by a distance of odd multiple of half

the wave length move with same speed and but in opposite direction. 
The minimum separation is $\frac{\lambda }{2}$
Option C.

A kite flying at a height h meter has r meter of string paid out at a time of t sec . If the kite moves horizontally with constant velocity v meter/sec then the at which the string is paid out is

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. $\sqrt{\left ( r^2 h^2 \right )v}$ mt/sec

  2. $\dfrac{v\sqrt{\left ( r^2 h^2 \right )}}{r}$ mt/sec

  3. $\dfrac{r\sqrt{\left ( r^2 h^2 \right )}}{v}$ mt/sec

  4. $\dfrac{\sqrt{\left ( r^2 h^2 \right )}}{rv}$ mt/sec

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Correct Option: C