Tag: wave velocity

Questions Related to wave velocity

 A solid cylinder of mg 50 kg and radius 0.5 m is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to and hanging freely. Tension in the string required to produce angular acceleration of revolutions $s ^ { - 2 }$ is

  1. 78.5 N

  2. 157 N

  3. 25 N

  4. 50 N


Correct Option: B
Explanation:

Mass=m=50kg

Radius=0.5m
Angular acceleration$=\alpha=2 rev/s^{2}$
Torque$=T \times R = T \times 0.5= \cfrac{T}{2} Nm ------(i)$
We know, $T=1 \alpha------(ii)$
From (i) and (ii),
$\cfrac{T}{2}=1 \alpha = \left( \cfrac{MR^{2}}{2} \right) \times (2 \times 2 \pi) rad/s^{2}$
$\therefore 1$ solid cylinder$=\cfrac{MR^{2}}{2}$
$\cfrac { T }{ 2 } =\cfrac { 50\times { (0.5) }^{ 2 } }{ 2 } \times 4\pi =50\pi =157N$

The vibration of a string of length 60 cm fixed at both ends are represented by $ y=4sin (\frac { \pi x}{15}) cos (96 \pi t) $ where x and y are in cm and t in second. the particle velocity at x=7.5 cm and t=0.25 s is

  1. Zero

  2. $ 10 cm s^{-1} $

  3. $ 100 cm s^{-1} $

  4. $ (4 \times 96) cm s^{-1} $


Correct Option: A

A $100$ Hz sinusoidal wave is travelling in the positive x-direction along a string with a linear mass density of $3.5 \times 10^{-3}$ kg/m and a tension of $35$ N. At time t = 0, the point x = 0 has zero displacements and the slope of the string is $\pi/20$. Then select the wrong alternative

  1. Velocity of wave is $100$ m/s

  2. Angular frequency is $(200 \pi)$ rad /s

  3. Amplitude of wave is $0.025$ m

  4. Propagation constant is $(4 \pi)$ $m^{-1}$


Correct Option: D

A uniform string fixed at both ends is vibrating in 3rd harmonic and equation $y = 4 ( \mathrm { cm } )$ $\sin \left[ \left( 0.8 \mathrm { cm } ^ { - 1 } \right) \times \right] \cos \left[ \left( 400 \pi \mathrm { s } ^ { - 1 } \right) t \right]$The length of the vibrating string is

  1. $6.75 \mathrm { m }$

  2. $12.45 \mathrm { m }$

  3. $11.8 \mathrm { m }$

  4. $18.7 \mathrm { m }$


Correct Option: C
Explanation:

$ y=4\sin  \left( { 0.8x } \right) \cos  \left( { 400\pi t } \right)  \ w=400\pi =2\pi f \ f=200\, Hz \ v=\dfrac { w }{ k } =\dfrac { { 400\pi \times 100 } }{ { 0.8 } } m/s \ 3\cdot \dfrac { v }{ { 2l } } =f \ \Rightarrow 200=\dfrac { { 3\times 400\pi \times 100 } }{ { 2\times l\times 0.8 } }  \ \Rightarrow l=\dfrac { { 3\times 400\pi  } }{ { 4\times 0.8 } }  \ =\dfrac { { 300\pi  } }{ { 0.8 } } \, cm \ =11.8\, m$

Hence,
option $(C)$ is correct answer.

The wave function for the wave pulse is $ Y (X,t) = \frac {0.1a^3}{a^2 +(X-Vt)^2}  with a = 4 cm. At X = 0 $ The displacement y (x,t) is observed to decreases from its maximum value to half of that value in time $ t = 2 \times 10^{-3} s $ choose the correct statement 

  1. The wave pulse is moving is negative X direction with speed 10 m/s

  2. The wave pulse is moving is positive X direction with speed 10 m/s

  3. The wave pulse is moving is negative X direction with speed 20 m/s

  4. The wave pulse is moving is positive X direction with speed 20 m/s


Correct Option: A

A string is properly tuned:

  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.


Correct Option: B

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

  1. $\sqrt { g h }$

  2. $\sqrt { g / h }$

  3. $\sqrt { 2 g h }$

  4. $\sqrt { h / g }$


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 

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

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

  3. $0$

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


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 :

  1. 200

  2. 80

  3. 300

  4. 180


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

  1. One second

  2. $0.5$ second

  3. $2\ seconds$

  4. Data given is insufficient


Correct Option: B