Tag: wave motion

Quizzes (25) Questions (122)

Questions Related to wave motion

Multiple choice physics wave motion reflection of waves

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  

  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. $Y\, =\,0.8 \,A\, sin\,(kx\,-\,\omega\,t\,+\,30^{\circ})$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice physics wave motion reflection of waves

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 :

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

Multiple choice physics wave motion reflection of waves

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  $x=$

  1. $4m$
  2. $3.75m$
  3. $3.50m$
  4. $3.25m$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $3+1=4\ m.$

Multiple choice physics wave motion reflection of waves

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 ($v = 340 m/s$)

  1. $25 cm$
  2. $6.25 cm$
  3. $62.5 cm$
  4. $2.5 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\dfrac{\lambda}{4}=6.25cm$

Multiple choice physics wave motion reflection of waves

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

  1. (0 m, 2 m)

  2. (0.5 m, 1.5 m)

  3. (0 m, 1.5 m)

  4. (0.5 m, 2 m)
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

 

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

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 $y=0.1m$>

  1. $1.9ms$
  2. $3.9ms$
  3. $2.4ms$
  4. $0.5ms$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

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).

  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. $y=A\cos kx \cos \omega t$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

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

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

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? 

  1. $400,600,800$
  2. $300,400,500$
  3. $100,150,200$
  4. $200,250,300$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given,  The Lowest frequency is $200cps$

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

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


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

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

  1. 211 HZ

  2. 199 Hz

  3. 208 Hz

  4. 202 Hz
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The initial frequency is 205 Hz. When tightened, the string produces 6 beats in 2 seconds, meaning the beat frequency is 6 / 2 = 3 Hz. Tightening a string increases its tension and thus its frequency. Therefore, the new frequency must be 205 + 3 = 208 Hz.