Physics

Wave Motion

536 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 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 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 optics interference

Light waves of wave length $\lambda$ propagate in a medium. If $M$ and $N$ are two points on the wave front and they are separated by a distance $\lambda /4$, the phase difference between them will be (in radian)

  1. $\dfrac{\pi}{2}$
  2. $\dfrac{\pi}{8}$
  3. $\dfrac{\pi}{45}$

  4. zero
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Points on the same wavefront are in the same phase. Therefore, the phase difference between any two points on the same wavefront is zero, regardless of the distance between them.

Multiple choice physics wave optics interference

For interference between waves from two sources of intensities $I$ and $4I$, find the intensity at the point in the pattern where the phase difference is $\dfrac{\pi}{2}$ and $\pi$.

  1. $10I$ and $I$
  2. $5I$ and $5I$
  3. $5I$ and $I$
  4. $5I$ and $10I$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

I_res = I1 + I2 + 2 * sqrt(I1 * I2) * cos(phi). I1 = I, I2 = 4I. For phi = pi/2, cos(pi/2) = 0, so I_res = I + 4I = 5I. For phi = pi, cos(pi) = -1, so I_res = I + 4I + 2 * sqrt(4 * I^2) * (-1) = 5I - 4I = I.

Multiple choice physics wave optics interference

The phase difference between two waves, represented by
${ y } _{ 1 }={ 10 }^{ -6 }sin{ 100t+(x/50)+0.5} m$
${ y } _{ 2 }={ 10 }^{ -6 }cos{ 100t+\left( \frac { x }{ 50 }  \right) } m$
where x is expressed in meters and is expressed in seconds, is approximately:

  1. 2.07 Radians

  2. 0.5 Radians

  3. 1.5 Radians

  4. 1.07 Radians
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

y1 = 10^-6 sin(100t + x/50 + 0.5). y2 = 10^-6 cos(100t + x/50) = 10^-6 sin(100t + x/50 + pi/2). The phase of y1 is phi1 = 100t + x/50 + 0.5. The phase of y2 is phi2 = 100t + x/50 + pi/2. Phase difference = |phi2 - phi1| = |pi/2 - 0.5| = |1.57 - 0.5| = 1.07 radians.

Multiple choice physics wave optics interference

What is the amplitude of resultant wave, when two waves $y _1=A _1\sin (\omega t-B _1)$ and $y _2=A _2\sin (\omega t-B _2)$ superimpose ?

  1. $A _1+A _2$
  2. $|A _1-A _2|$
  3. $\sqrt{A _1^2+A _2^2+2A _1A _2\cos (B _1-B _2)}$
  4. $\sqrt{A _1^2+A _2^2+2A _1A _2\cos B _1 B _2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When two waves y1 = A1 sin(omega*t - B1) and y2 = A2 sin(omega*t - B2) superimpose, the resultant amplitude A is given by the vector sum of the amplitudes: A = sqrt(A1^2 + A2^2 + 2*A1*A2*cos(B1 - B2)).

Multiple choice physics superposition and interference of sound waves

Two sources of sound A and B produce the wave of $350Hz$, they vibrate in the same phase. The particle $P$ is vibrating under the influence of these two waves. If the amplitude at the point $P$ produced by the two waves is $0.3mm$ and $0.4mm$ then the resultant amplitude of the point $P$ will be: (path difference $AP-BP=25cm$ and the velocity of sound is $350m/sec$)

  1. $0.7mm$
  2. $0.1mm$
  3. $0.2mm$
  4. $0.5mm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\begin{array}{l} \lambda =\dfrac { v }{ n } =\dfrac { { 350 } }{ { 350 } } =1m=100cm \ Path\, difference\, \, \Delta x=AP-BP=25cm \ Phase\, difference\, \, \Delta \varphi =\dfrac { { 2\pi  } }{ \lambda  } \Delta x=\dfrac { { 2\pi  } }{ 1 } \times \left( { \dfrac { { 25 } }{ { 100 } }  } \right) =\dfrac { \pi  }{ 2 }  \ Amplitude\, A=\sqrt { { { \left( { { a _{ 1 } } } \right)  }^{ 2 } }+{ { \left( { { a _{ 2 } } } \right)  }^{ 2 } } } =\sqrt { { { \left( { 0.3 } \right)  }^{ 2 } }+{ { \left( { 0.4 } \right)  }^{ 2 } } } =0.5mm \end{array}$

Multiple choice physics superposition and interference of sound waves

 When two sound waves with a phase of $\dfrac { \pi  }{ 2 } $ and each having amplitude A and frequency $\omega $, are superimposed on each other, then the maximum amplitude and frequency  of resultant wave is: 

  1. $\sqrt { 2 } A;\omega $
  2. $\dfrac { A }{ \sqrt { 2 } } ;\dfrac { \omega }{ 2 } $
  3. $\left( \sqrt { 2 } \right) A;\dfrac { \omega }{ 2 } $
  4. $\dfrac { A }{ \sqrt { 2 } } ;\omega $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The resultant amplitude of two waves with amplitude A and phase difference phi is sqrt(A^2 + A^2 + 2*A*A*cos(phi)). For phi = pi/2, this is sqrt(2*A^2) = A*sqrt(2). The frequency remains the same.

Multiple choice physics superposition and interference of sound waves

Two waves having the intensities in the ratio 9 : 1 produce interference. The ratio of maximum to minimum intensity is equal to

  1. 4 : 1

  2. 9 : 1

  3. 2 : 1

  4. 10 : 8
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the intensity of the two waves be $I _1$  and  $I _2$


Given:    $I _2  :  I _1  =  9  :  1        \implies  I _2   =  9  I _1$

Now       $\dfrac{I _{max}}{I _{min}}  =  \dfrac{(\sqrt{I _1} + \sqrt{I _2})^2}{(\sqrt{I _2} - \sqrt{I _1})^2} = \dfrac{(\sqrt{I _1} + \sqrt{9  I _1})^2}{(\sqrt{9  I _1} - \sqrt{I _1})^2} = \dfrac{16  I _1}{4  I _1}$ 

$\implies    I _{max}  :  I _{min}  =  4  : 1$

Multiple choice physics superposition and interference of sound waves

Two waves $Y _{1}= asin\omega t$  and $Y _{2}= asin(\omega t+\delta )$  are  producing interference, then resultent intensity is 

  1. $a^{2}cos^{2}\delta /2$
  2. $2a^{2}cos^{2}\delta /2$
  3. $3a^{2}cos^{2}\delta /2$
  4. $4a^{2}cos^{2}\delta /2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics superposition and interference of sound waves

When interference is produced by two progressive waves of equal frequencies, then the maximum intensity of the resulting sound are N times the intensity of each of the component waves. The value of N is

  1. 1

  2. 2

  3. 4

  4. 8
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$y _1=A _0 \sin{\omega t}$, 

$y _2=A _0 \sin{\omega t+ \phi}$,

also, $I \propto y^2={(y _1+y _2)}^2={(2A _0 \sin{\omega t +\phi/2} \cos{\phi/2})}^2=4{A _0}^2{(\sin{\omega t +\phi/2})}^2{(\cos{\phi/2})}^2$,

Multiple choice physics superposition and interference of sound waves

In case of super position of waves (at $x=0$),
 $y _{1}=4\sin(1026\pi t)$ and $y _{2}=2\sin(1014\pi t)$


a) the frequency of resulting wave is $510$ Hz
b) the amplitude of resulting wave varies at the frequency of $3$ Hz
c) the frequency of beats is $6$ per second
d) the ratio of maximum to minimum intensity is $9$

The correct statements are


  1. a,d only

  2. b,d only

  3. a, c, d only

  4. a,b,c,d
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Beat $=\delta _{1}-\delta _{2}$$=\dfrac{\omega _{1}}{2\pi}-\dfrac{\omega _{2}}{2\pi}$$=\dfrac{1026\pi}{2\pi}-\dfrac{1014\pi}{2\pi}$$=6$

$\dfrac{I max}{I min}=\dfrac{(\Delta _{1}+A _{2})^{2}}{(\Delta _{1}A _{2})^{2}}=\dfrac{(4+2)^{2}}{(4-2)^{2}}=\dfrac{36}{4}=\dfrac{9}{1}$

$y _{1}=4 sin (1026 \pi t)$
$y _{2}=2 sin (1014 \pi t)$
$y=y _{1}+y _{2}$
   $=4 Sin (1026 \pi t)+2 sin (1014\pi t)$
   $=\left(4\sqrt{(\dfrac{3}{1})^{2}+cos (12\pi t)}\right ) sin (1020 \pi t)$

So, clearly frequency $ =\dfrac{\omega}{2\pi}=\dfrac{1020\pi}{2\pi}=510 Hz$
and amplitude of resulting wave varies at frequency
$\delta=\dfrac{\omega}{2\pi}=\dfrac{6\pi}{2\pi}=3Hz$

Multiple choice standing waves waves physics

Which of the following function represent traveling waves?

  1. $ y = (x + 5t)^3 $
  2. $ y = tan (2x + 3t) $
  3. $ y = \theta^{(4t+2x)^2} $
  4. $ y = \frac { 1 }{ x + 3t } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A function represents a traveling wave if it can be written in the form f(ax +/- bt). Option D, y = 1/(x + 3t), fits this form, representing a wave traveling in the negative x-direction.

Multiple choice standing waves waves physics

A progressive wave is incident normally on a flat reflector. The reflected wave overlaps with the incident wave and a stationary wave is formed.
At an antinode, what could be the ratio $\dfrac{displacement of the incident wave}{displacement of the reflected wave}$ at any instant?

  1. $-1$
  2. $0$
  3. $1$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Standing wave is formed when two waves with equal magnitude and and opposite direction overlap Both waves have same amplitude at antinode and opposite amplitude at node.

Multiple choice standing waves waves physics

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 X = $\dfrac { v\lambda }{ 2 } $;n = 0,1,2,..................
  2. A standing wave have nodes at X=$\left( n+\dfrac { 1 }{ 2 } \right) \dfrac { \lambda }{ 2 } $;n = 0,1,2,...............................

  3. A wave travelling along +x direction.

  4. A wave travelling along -x direction
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Intensity wave is standing wave and have nodes at 
$X=\left( n+\dfrac { 1 }{ 2 }  \right) \dfrac { \lambda  }{ 2 } $ n=0,1,2,......