Tag: wave velocity

Questions Related to wave velocity

Which of the following equations represents a transverse wave travelling along -y axis?

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

  1. $x = A\sin\ (\omega t\ -\ ky)$

  2. $x= A\ sin\ (\omega t\ +\ ky)$

  3. ${ y } _{ 0 }\ =A\sin\ (\omega t - kX   )$

  4. ${ y } _{ 0 } = A\ sin (\omega t + kX )$

Show answer
Correct Option: B
Explanation:

$\begin{array}{l} For\, \, negative\, \, y-axis \ sign\, \, of\, \, \omega t\, \, & \, \, ky\, \, should\, \, be\, \, same\,  \ x=A\sin  \left( { \omega t+ky } \right)  \ Hence, \ option\, \, B\, \, is\, correct\, \, naswer. \end{array}$

The displacement from the position of equilibrium of a point $4\ cm$ from a source of sinusoidal oscillations is half the amplitude at the moment $t=\dfrac{T}{6} (T$ is the time period$)$. Assume that the source was at mean position at $t=0$. The wavelength of the running wave is 

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

  1. $0.96\ m$

  2. $0.48\ m$

  3. $0.24\ m$

  4. $0.12\ m$

Show answer
Correct Option: B
Explanation:

Going by the data given to us, this wave is sinusoidal in nature and the wave equation takes the form of
$y = A\sin( \omega t - kx),$ as it is given that at $t = 0,$  the source is at mean position.
Here$,\ x = 4\ cm = 0.04\ m$
$y = A/2$
Amplitude $= A$
$t = \dfrac{T}{6}$
We know that $ \omega  = 2 \dfrac{ \pi }{T}$
$\Rightarrow \dfrac{A}{2} = A\sin((2  \pi  / T)(T/6) - 0.04k)$
$\sin((2 \pi  / T)(T/6) - 0.04k) = 1/2$
$\Rightarrow ((2  \pi / T)(T/6) - 0.04k) =  \pi / 6$
$k =  \pi  / 0.24$
wavelength $ \lambda = 2\pi  / k = 0.48\ m$

A string of length 1 m fixed at one end and on the other end a block of mass M=4 kg is suspended.The string is set into vibrations and represented by equation, Y=$6\sin \left( {\dfrac{{\pi x}}{{10}}} \right)\;\cos \;100\;\pi t,$  where x and y are in cm an in seconds.
Find the number of loops formed in the string.

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

  1. 3

  2. 4

  3. 5

  4. 6

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Correct Option: A

A travelling wave is given by $y=\frac { 0.8 }{ 3{ x }^{ 2 }+12xt+12{ t }^{ 2 }+1 } $ where x and y are is m and t is in sec, then velocity and amplitude wave will be

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

  1. 2m/s, 0.2m

  2. 4m/s, 0.2m

  3. 2m/s, 0.4m

  4. none

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Correct Option: D

A travelling wave on a light on a tight string is described by the equation $y=A\sin (kx-\omega t)$. if tension in the string is $F$ then total energy stored in the string having from $x=0$ to $x=2\pi/k$ is 

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

  1. $\pi FA^{2}$

  2. $\pi kFA^{2}$

  3. $\pi k^{2}FA$

  4. $none\ of\ these$

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Correct Option: A

The $(x, y)$ co-ordinates of the corners of a square plate are $(0, 0) (L, 0) (L, L)$ & $(0, L)$. The edges of the plate are clamped & transverse standing waves are set up in it. If $u (x, y)$ denotes the displacement of the plate at the point $(x, y)$ at some instant of time, the possible expression(s) for $u$ is/are : ($a$ = positive constant) 

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

  1. $a\displaystyle \cos \left(\dfrac{\pi x}{2 L}\right)$ $\displaystyle \cos \left(\dfrac{\pi y}{2 L}\right)$

  2.  $a\displaystyle \sin \left(\dfrac{\pi x}{L}\right)$ $\displaystyle \sin \left(\dfrac{\pi y}{L}\right)$

  3. $a\displaystyle \sin \left(\dfrac{\pi x}{L}\right)$ $\displaystyle \sin \left(\dfrac{2\pi y}{L}\right)$

  4.  $a\displaystyle \cos \left(\dfrac{2\pi x}{L}\right)$ $\displaystyle \sin \left(\dfrac{\pi y}{L}\right)$

Show answer
Correct Option: B,C
Explanation:
The expression for $u(x,y)$ should satisfy the following conditions-
i) $u=0$ at $x=0$ and at $y=0$
ii) $u=0$ at $x=L$ and at $y=L$
Only choices B and C satisfy this condition.

The displacement of the particle at $x=0$ of a stretched string carrying wave in the positive x-direction is given $f(t)=A sin \frac {t} {T})$. The wave speed is V. Write the wave equation 

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

  1. $f(x,t)=A sin (\frac {t} {T}) - (\frac{x} {V})$

  2. $f(x,t)=A sin (\frac {t} {T}) + (\frac{x} {VT})$

  3. $f(x,t)=A sin (t+- (\frac{x} {V})$

  4. $f(x,t)=A sin (\frac {t} {T}) - (\frac{x} {VT})$

Show answer
Correct Option: D

A uniform string of length $L$ fixed between the two ends is vibrating in three segments. The wavelength of wave in string is

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

  1. $\dfrac { L }{ 3 } $

  2. $3L$

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

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

Show answer
Correct Option: C
Explanation:

$\begin{array}{l} \dfrac { { 3\lambda  } }{ 2 } =L \ \lambda =\dfrac { { 2l } }{ 3 }  \end{array}$

$\therefore $ Option $C$ is correct.

A uniform rope of length $L$ and mass ${m _1}$ hangs vertically from a rigid support. A block of mass ${m _{2\,}}$ is attached to the free end of the rope. A transverse pulse of wavelength ${\lambda _1}$ is produced at the lower end of the rope. The Wavelength of the pulse when it reaches the top of the rope is ${\lambda _2}$. The ratio ${\lambda _2}/{\lambda _1}$ is 

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

  1. $\sqrt {\frac{{{m _1} + {m _2}}}{{{m _1}}}} $

  2. $\sqrt {\frac{{{m _1}}}{{{m _2}}}} $

  3. $\sqrt {\frac{{{m _1} + {m _2}}}{{{m _2}}}} $

  4. $\sqrt {\frac{{{m _2}}}{{{m _1}}}} $

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Correct Option: C

A stretched string of length $1m$ fixed at both ends, having a mass of $5\times{10}^{-4}kg$ is under a tension of $20N$. It is plucked at a point situated at $200cm$ from one end. The stretched string would vibrate with a frequency of

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

  1. $200Hz$

  2. $100Hz$

  3. $250Hz$

  4. $256Hz$

Show answer
Correct Option: A
Explanation:

$\begin{array}{l} v=\dfrac { 2 }{ { 2l } } \sqrt { \dfrac { { T\times l } }{ M }  }  \ or\, \, ,\, \, \sqrt { \dfrac { T }{ { Ml } }  } =\sqrt { \dfrac { { 20 } }{ { 5\times 10-4\times l } }  }  \ =\sqrt { 4\times { { 10 }^{ 4 } } } Hz=200Hz \end{array}$