Tag: wave motion

Quizzes (25) Questions (122)

Questions Related to wave motion

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

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.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Beats are produced when two sound waves of slightly different frequencies interfere. When the two frequencies are identical, the beat frequency becomes zero, meaning the beat frequency vanishes, indicating that the string is properly tuned.

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

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. $\sqrt { h / g }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The tension at a distance h from the free end of a hanging rope is T = mu * g * h, where mu is the linear mass density. The wave speed is v = sqrt(T/mu) = sqrt(mu * g * h / mu) = sqrt(g * h).

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Linear mass density mu = mass / length = 2.5 kg / 20 m = 0.125 kg/m. Wave speed v = sqrt(T / mu) = sqrt(200 / 0.125) = sqrt(1600) = 40 m/s. Time taken t = distance / speed = 20 m / 40 m/s = 0.5 s.

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

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 

  1. 62.8 cm / s

  2. 75 cm / s

  3. 100 cm / s

  4. 121 cm / s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The wave equation is y = 10 sin(0.01 * pi * x - 2 * pi * t). The transverse velocity is v_y = dy/dt = -10 * 2 * pi * cos(0.01 * pi * x - 2 * pi * t). The maximum transverse speed is 20 * pi = 20 * 3.14 = 62.8 cm/s.

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

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 :

  1. $0$
  2. $\dfrac{5 \pi}{6}$
  3. $\pi$
  4. $\dfrac{5 \pi}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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$\pi$

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

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

  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. $7.5 \mathrm { cm } , 75 \mathrm { cm } \mathrm { s } ^ { - 1 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 .

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

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

  1. $\lambda / 4$
  2. $\lambda / 3$
  3. $\lambda / 2$
  4. $\lambda $
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer