Tag: wave motion

Questions Related to wave motion

A wave travels on a light string. The equation of the waves is $Y\, = \,A\, sin\,(kx\,-\,\omega\,t+\,30^{\circ})$. It is reflected from a heavy string tied to end of the light string at x = 0 . If 64% of the incident energy is reflected then the equation of the reflected wave is  

physics wave motion reflection of waves

  1. $Y\, =\,0.8 \,A\, sin\,(kx\,-\,\omega\,t\,+\,30^{\circ}\,+\,180^{\circ})$

  2. $Y\, =\,0.8 \,A\, sin\,(kx\,+\,\omega\,t\,+\,30^{\circ}\,+\,180^{\circ})$

  3. $Y\, =\,0.8 \,A\, sin\,(kx\,-\,\omega\,t\,-\,30^{\circ})$

  4. $Y\, =\,0.8 \,A\, sin\,(kx\,-\,\omega\,t\,+\,30^{\circ})$

Show answer
Correct Option: B
Explanation:

There are three things we need to take into account:

  • Energy transfer
  • Change in velocity
  • Change in phase
We know that power delivered is proportional to $A^{2}$
Hence if power(energy) reduces to 64%. We get that Amplitude must reduce to 80% or 0.8A.
Now the reflected wave is moving in the opposite direction. (velocity is negative now).
Also because of the hard soft boundary reflection (there is a phase lag of $180^{\circ}$
Hence the new equation becomes:
$y = 0.8A sin(kx + \omega t + 30^{\circ} + 180^{\circ})$
Hence option B.

A pulse of a wave train travels along a stretched string and reaches the fixed end of the string. It will be reflected back with :

physics wave motion reflection of waves

  1. a phase change of ${180}^{o}$ with velocity reversed

  2. the same phase as the incident pulse with no reversal of velocity

  3. a phase change of ${180}^{o}$ with no reversal of velocity

  4. the same phase as the incident pulse but with velocity reversed

Show answer
Correct Option: A
Explanation:

A pulse of wave train when travels along a stretched string and reaches the fixed end of the string, then it will be reflected back to the same medium and the reflected ray suffers a phase change of $\pi$ with the incident wave but wave velocity after reflection reverses its direction.

A wave of length $2m$ is superposed on its reflected wave to form a stationary wave. A node is located at  $ x=3m$ The next node will be located at  $x=$

physics wave motion reflection of waves

  1. $4m$

  2. $3.75m$

  3. $3.50m$

  4. $3.25m$

Show answer
Correct Option: A
Explanation:

Since wave length is $2m$, half of wavelength is $1m$. Node forms after each half of wave length. So, node will be formed at each $1 m$
So, as node is formed at $3 m$
next node will be formed at $3+1=4\ m.$

A sound wave of frequency $1360 Hz$ falls normally on a perfectly reflecting  wall.  The shortest distance from the wall at which the air particles have maximum amplitude of vibration is ($v = 340 m/s$)

physics wave motion reflection of waves

  1. $25 cm$

  2. $6.25 cm$

  3. $62.5 cm$

  4. $2.5 cm$

Show answer
Correct Option: B
Explanation:

$v=f\lambda$
$\lambda=\dfrac{340}{1360}=\dfrac{1}{4}=0.25m$
$=25cm$
As the end is a node, shortest distance at which antinode is formed is $\dfrac{\lambda}{4}=6.25cm$

A string fixed at one end only is vibrating in its third harmonic. The wave function is $y(x,t) = 0.02 sin(3.13x) cos(512t)$, where y and x are in metres and t is in seconds. The nodes are formed at positions

physics wave motion reflection of waves

  1. (0 m, 2 m)

  2. (0.5 m, 1.5 m)

  3. (0 m, 1.5 m)

  4. (0.5 m, 2 m)

Show answer
Correct Option: B
Explanation:

Node in formed only at the finest end of the string and the free end acts as an anti node.

In the third harmonics two nodes are formed:

$y\left( x,t \right) =0.02\sin { \left( 3.13x \right) \cos { \left( 512t \right)  }  } $

Standard equation given by

$y\left( x,t \right) =2a\sin { \left( \cfrac { 2\pi  }{ \lambda  } x \right)  } \cos { \left( 2\pi vt \right)  } $

Comparing both equation, we get

$3.13x\quad =\cfrac { 2\pi  }{ \lambda  } x\\ or,\quad \lambda =\cfrac { 2\lambda  }{ 3.13 } \\ \quad \quad \quad \quad =2 m (approx)$

The nodes are formed at $\cfrac { \lambda  }{ 4 } =0.5$ from origin and at $\cfrac { 3\lambda  }{ 4 } =1.5$ from origin.

 

A transverse wave on a string has an amplitude of $02m$ and a frequency of $175Hz$. Consider a particle of the string at $x=0$. It begins with a displacement $y=0$ at $t=0$, according to equation $y=0.2\sin{(kx+\omega t)}$. How much time passes between the first two instant when this particle has a displacement of $y=0.1m$>

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

  1. $1.9ms$

  2. $3.9ms$

  3. $2.4ms$

  4. $0.5ms$

Show answer
Correct Option: C

For a string clamped at both its ends, which of the following wave equation is/are valid for a stationary wave set up in it? (Origin is at one end of string).

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

  1. $y=A\sin kx.\sin \omega t$

  2. $y=A\cos kx \sin \omega t$

  3. $y=A\sin kx. \cos \omega t$

  4. $y=A\cos kx \cos \omega t$

Show answer
Correct Option: A,C
Explanation:

For all values of t, y$=0$ at $x=0$
Hence, (A) and (C) are correct.

In transverse wave the distance between a crest and through at the same place is 1.0 cm. The next crest appears at the same place after a time interval of 0.4 s. The maximum speed of the vibrating particles in the same medium is :

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

  1. $\dfrac{3\pi }{2}cm/s$

  2. $\dfrac{5\pi }{2}cm/s$

  3. $\dfrac{\pi }{2}cm/s$

  4. $2\pi cm/s$

Show answer
Correct Option: A

A certain strings will resonate to several frequencies , the lowest of which is $200$cps.what are the next three higher frequencies to which it resonates? 

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

  1. $400,600,800$

  2. $300,400,500$

  3. $100,150,200$

  4. $200,250,300$

Show answer
Correct Option: A
Explanation:
Given,  The Lowest frequency is $200cps$

Let  $f$ resonant the fundamental frequency, then the next higher frequency is: $2f,3f,4f$

$2\times200=400cps,3\times200=600,4\times200=800cps$


The string of a violin emits a note of 205 Hz at its correct tension. The string is tightened slightly and then it produces six beats in two seconds with a tuning fork of frequency 205 Hz. The frequency of the note emitted by the taut string is

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

  1. 211 HZ

  2. 199 Hz

  3. 208 Hz

  4. 202 Hz

Show answer
Correct Option: A