Physics

Wave Motion

489 Questions

Wave motion questions cover the principles of traveling and stationary waves, including their equations and intensities. The topics explore interference patterns, phase differences, and electromagnetic radiation speeds. Mastery of these concepts is vital for physics sections in engineering and civil services examinations.

Wave interferenceStanding wavesPhase differenceElectromagnetic radiationWave equations

Wave Motion Questions

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

Two waves of intensities $I$ and $4I$ superimpose. The minimum and maximum intensities will respectively be

  1. $I,\space 9I$
  2. $3I,\space 5I$
  3. $I,\space 5I$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The Intensity of the wave is directly proportional to the square of its amplitude.
$I \propto A^{2}$;
$I = cA$$^2$;   '$c$' is an arbitrary constant.
So, if a wave with Intensity $I$ has an amplitude of $A(A{ _{1}}$)
A wave with an Intensity of $4I$ would respectively have an amplitude of $2A$($A{ _{2}}$)

If 2 waves with amplitudes $A{ _{1}}$,$A{ _{2}}$ are superimposed, the resultants would be
Maximum of $A{ _{1}}$ + $A{ _{2}} = 3A$ (Constructive Interference)
Minimum of $|A{ _{1}}$ - $A{ _{2}} |   = A $ (Destructive Interference)
The wave of amplitude $3A$ would have an Intensity of $9I$
The wave of amplitude $A$ would have an Intensity of $I$
Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

For a wave displacement amplitude is $10^{-8} m$ density of air $1.3 kg m^{-3}$ velocity in air $340 ms^{-1}$ and frequency is 2000 Hz.The average intensity of wave is

  1. $5.3\times 10^{-4} Wm^{-2}$
  2. $5.3\times 10^{-6} Wm^{-2}$
  3. $3.5\times 10^{-8} Wm^{-2}$
  4. $3.5\times 10^{-6} Wm^{-2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Intensity I = 2 * pi^2 * f^2 * A^2 * rho * v. Plugging in values: I = 2 * (3.14)^2 * (2000)^2 * (10^-8)^2 * 1.3 * 340. I = 2 * 9.86 * 4 * 10^6 * 10^-16 * 1.3 * 340 = 5.3 * 10^-4 W/m^2.

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

Two sound waves of equal intensity $I$ superimpose at point $P$ in $90^{\small\circ}$ out of phase. The resultant intensity at point $P$ will be

  1. $4I$
  2. $\sqrt2I$
  3. $2I$
  4. $I$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Amplitude of the resultant wave is:
$A _R=\sqrt{A^2 _1+A^2 _2+2A _1A _2\cos(\theta)}$.

Here, $A _1=A _2=A \text{ and } \theta = \pi/2$

So, $A _R=A\sqrt{2(1+\cos(\theta))}=2A\cos(\theta/2)=2A\cos(\pi/4)=\sqrt{2}A$ 

$\Rightarrow I _R=|A _R|^2=2A^2=2I$

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

A wave of frequency 500$\mathrm { Hz }$ travels between $\mathrm { x }$and $\mathrm { Y }$ and travel a distance of 600$\mathrm { m }$ in 2$\mathrm { sec }$ . between $X$ and $Y .$ How many wavelength are therein distance $X Y$ :

  1. 1000

  2. 300

  3. 180

  4. 2000
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Velocity v = distance/time = 600m / 2s = 300 m/s. Wavelength lambda = v/f = 300/500 = 0.6 m. Number of wavelengths = distance / lambda = 600 / 0.6 = 1000.

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

Four independent waves are represented by the equations :
$y _1 = a _1\  sin\  \omega t$
$y _2 = a _2\ sin\  \omega t$
$y _3 = a _3\ cos\  \omega t$
$y _4 = a _4\ sin\  (\omega t + \pi/3)$ 
Then the waves for which phenomenon of interference will be observed are - 

  1. 1 and 3

  2. 1 and 4

  3. all 1, 2, 3 and 4

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Interference is observed between waves of the same frequency. Waves 1 (sin wt) and 3 (cos wt = sin(wt + pi/2)) have the same frequency (omega). Wave 2 also has frequency omega, but 1 and 3 are the standard pair for demonstrating interference.

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

Two sinusoidal plane waves same frequency having intensities $I _0 $ and $ 4I _0 $ are travelling in same direction. The resultant intensity at a point at which waves meet with a phase difference of zero radian is

  1. $ I _0$
  2. $5 I _0$
  3. $9 I _0$
  4. $3 I _0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let, $I _1=I _0  $ and $  I _2=4I _0 $

Resultant intensity, $I=I _1+I _2+2\sqrt{I _1I _2} cos\phi $
                                   $= I _0+4I _0+2\sqrt{I _04I _0} cos0^{\circ} \ =9I _0 $

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

If the ratio of maximum to minimum intensity in beat is 49, then the ratio of amplitudes of two progressive wave trains

  1. 7:1

  2. 4:3

  3. 49:1

  4. 16:9
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\dfrac{I _{max}}{I _{min}}=\dfrac{(\sqrt{I _1}+\sqrt{I _2})^2}{(\sqrt{I _1}-\sqrt{I _2})^2}=49$

$\dfrac{(\sqrt{I _1}+\sqrt{I _2})}{(\sqrt{I _1}-\sqrt{I _2})}=7$

$\sqrt{I _1}+\sqrt{I _2}=7(\sqrt{I _1}-\sqrt{I _2})$

$\dfrac{\sqrt{I _1}}{\sqrt{I _2}}=\dfrac{4}{3}$

$\dfrac{a}{b}=\dfrac{4}{3}$

Here, $a =\sqrt{I _1}$ and $b =\sqrt{I _2}$, where a and b are the amplitudes of the two progressive waves.

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

If the phase difference between two sound waves of wavelength $  \lambda  $ is $  60^{\circ} $, the corresponding path difference is

  1. $ \frac{\lambda}{6} $
  2. $ \frac{\lambda}{2} $
  3. $ \lambda 2 $
  4. $ \frac{\lambda}{4} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The relationship between phase difference (delta phi) and path difference (delta x) is delta phi = (2 * pi / lambda) * delta x. Given delta phi = 60 degrees = pi/3 radians, pi/3 = (2 * pi / lambda) * delta x. Solving for delta x gives delta x = lambda / 6.

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

Equations of stationary and a travelling wave are as follows: $Y _1=sin\, kx\, cos\,\omega t$ and $Y _2=a\, sin\, (\omega t-kx)$. The phase difference between two points $X _1=\dfrac{\pi}{3k}$ and $ X _2=\dfrac{3\pi}{2k}$ are $\phi _1$ and $\phi _2$ respectively for the two waves.The ratio of $\dfrac{\phi _1}{\phi _2}$ is 

  1. 6

  2. 5

  3. 4

  4. 2
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

Two waves of intensities 1 and 4 superimposes. Then the maximum and minimum intensities are :

  1. 9 and 1

  2. 31 and 1

  3. 91 and 31

  4. 61 and 1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Ratio of amplitudes $\sqrt{\dfrac{4}{1}}=\dfrac{2}{1}$
$\dfrac{maximum\ amplitude}{minimum\ amplitude}=\dfrac{2+1}{2-1}=\dfrac{3}{1}$


$\dfrac{maximum\ intensity}{minimum\ intensity}=\left (\dfrac{3}{1}  \right )^2=\dfrac{9}{1}$

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

Two periodic waves of intensities ${I} _{1}$ and ${I} _{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities possible is :

  1. ${I} _{1} + {I} _{2}$
  2. ${\left(\sqrt{{I} _{1}} + \sqrt{{I} _{2}}\right)}^{2}$
  3. ${\left(\sqrt{{I} _{1}} - \sqrt{{I} _{2}}\right)}^{2}$
  4. $2\left({I} _{1} + {I} _{2}\right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Resultant intensity of two periodic waves is given by
$I={ I } _{ 1 }+{ I } _{ 2 }+2\sqrt { { I } _{ 1 }{ I } _{ 2 }\cos { \delta  }  } $
where $\delta$ is the phase difference between the waves.
For maximum intensity,
$\delta =2n\pi ;n=0,1,2,...$etc.
Therefore, for zero order maxima, $\cos { \delta  } =1$
${ I } _{ max }={ I } _{ 1 }+{ I } _{ 2 }+2\sqrt { { I } _{ 1 }{ I } _{ 2 } } ={ \left( \sqrt { { I } _{ 1 } } +\sqrt { { I } _{ 2 } }  \right)  }^{ 2 }$
For minimum intensity,
$\delta =\left( 2n-1 \right) \pi ;n=1,2,...$etc.
Therefore, for Ist order minima, $\cos { \delta  } =-1$
${ I } _{ min }={ I } _{ 1 }+{ I } _{ 2 }-2\sqrt { { I } _{ 1 }{ I } _{ 2 } } $
$={ \left( \sqrt { { I } _{ 1 } } -\sqrt { { I } _{ 2 } }  \right)  }^{ 2 }$
Therefore,  ${ I } _{ max }+{ I } _{ min }={ \left( \sqrt { { I } _{ 1 } } +\sqrt { { I } _{ 2 } }  \right)  }^{ 2 }+{ \left( \sqrt { { I } _{ 1 } } -\sqrt { { I } _{ 2 } }  \right)  }^{ 2 }$
$=2\left( { I } _{ 1 }+{ I } _{ 2 } \right) $

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

A travelling wave represented by $y=A\sin { \left( \omega t-kx \right)  } $ is superimposed on another wave represented by $y=A\sin { \left( \omega t+kx \right)  } $. The resultant is 

  1. A standing wave having nodes at$\quad x=\left( n+\cfrac { 1 }{ 2 } \right) \cfrac { \lambda }{ 2 } $, where $n=0,1,2$
  2. A wave travelling along $+x$ direction
  3. /a wavelength travelling along $-x$ direction
  4. a standing wave having nodes at $x=\cfrac { n\lambda }{ 2 } $, where $n=0,1,2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to the principle of superposition, the resultant wave is
$y = asin(kx - \omega t) + asin(kx + \omega t)$
$= 2a\ sin\ \omega t\ cos\ x$                                                 .....(i)

It represents a standing wave.
In the standing wave, there will be nodes (where amplitude is zero) and antinodes  (where amplitude is largest).
From Eq. (i), the positions of nodes are given by
$sin\ kx = 0 \Longrightarrow kx = n\pi; n = 0, 1, 2, ....$

or $\dfrac{2\pi}{\lambda}x = n\pi; 0, 1, 2, ....$

or $x = \dfrac{n\lambda}{2}; n = 0, 1, 2, ...$
In the same way,
From Eq.(i), the positions of antinodes are given by$|sinkx| = 1$
$\Longrightarrow kx = (n + \dfrac{1}{2})\pi ; n = 0, 1, 2, ..... $

or $\dfrac{2\pi x}{\lambda} =  (n + \dfrac{1}{2})\pi ; n = 0, 1, 2, ..... $

or $x =  (n + \dfrac{1}{2})\dfrac{\lambda}{2} ; n = 0, 1, 2, ..... $

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

Consider the superposition of N harmonic waves of equal amplitude and frequency. If N is a very large number determine the resultant intensity in terms of the intensity $\left( { I } _{ 0 } \right)$ of each component wave for the condition when the component waves have identical phases.

  1. ${ NI } _{ 0 }$
  2. ${ N }^{ 2 }{ I } _{ 0 }$
  3. $\sqrt { N } { I } _{ 0 }$
  4. ${ I } _{ 0 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As all the waves are in phase and having same amplitude and frequency

So, the intensities will simply get add to give the resultant intensity
$\Rightarrow (Intensity) _{Resultant}=(I _0+I _0+.....I _0)-N\quad times\ \quad\quad\quad\quad\quad=NI _0$

Multiple choice physics stationary waves determining wavelength and speed of sound resonance tube resonance and sonometer

A sound wave of wavelength $\lambda$ travels towards the right horizontally with a velocity $V$. It strikes and reflects from a vertical plane surface, traveling at a speed $v$ towards the left. The number of positive crests striking in a time interval of $3s$ on the wall is:

  1. $3(V+v)/ \lambda$
  2. $3(V-v)/ \lambda$
  3. $(V+v)/3\lambda$
  4. $(V-v)/3\lambda$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The relative velocity of the wave crests with respect to the wall is (V + v). The distance covered in time t is (V + v) * t. The number of crests is distance / wavelength = (V + v) * t / lambda. For t = 3s, this is 3(V + v) / lambda.