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

The equation of a wave travelling on a string is $y=4 sin \left[ \dfrac { \pi  }{ 2 } \left( 8t-\dfrac { x }{ 8 }  \right)  \right] $, where $x,y$ are in cm and $t$ is in second. The velocity of the wave is

  1. $64 cm/s$ in $-x$ direction
  2. $32 cm/s$ in $-x-$ direction
  3. $32 cm/s$ in $+x-$ direction
  4. $64 cm/s$ in $+x-$ direction
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation is y = 4 sin(4 * pi * t - pi * x / 16). Comparing to y = A sin(omega * t - k * x), omega = 4 * pi and k = pi / 16. Wave speed v = omega / k = 4 * pi / (pi / 16) = 64 cm/s. Since the signs of omega * t and k * x are opposite, the wave travels in the +x direction.

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

Two travelling waves $y _1=A sin[k(x-ct)]$ and $y _2\, sin[k(x+ct)]$ are superimposed on string. The distance between adjacent nodes is 

  1. $c\, t/\pi$
  2. $c\, t/2\pi$
  3. $\pi/2k$
  4. $\pi/k$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The superposition of two waves traveling in opposite directions forms a standing wave. The distance between adjacent nodes is lambda / 2. Since k = 2 * pi / lambda, lambda / 2 = pi / k.

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

A wave propagates on a string in positive $x-$ direction with a speed of $40\ cm/s$. The shape of string at $t=2\ s$ is $y=10\cos \,\dfrac{x}{5}$, where $x$ and $y$ are in centimetre. The wave equation is :

  1. $y=10\cos \left(\dfrac{x}{5}-8t\right)$
  2. $y=10\sin \left(\dfrac{x}{5}-8t\right)$
  3. $y=10\cos \left(\dfrac{x}{5}-8t+16\right)$
  4. $y=10\sin \left(\dfrac{x}{5}-8t+16\right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Wave speed v = 40 cm/s. At t = 2 s, y = 10 cos(x / 5). The general form is y = 10 cos(x / 5 - omega * t + phi). Since v = omega / k, omega = v * k = 40 * (1 / 5) = 8 rad/s. At t = 2, y = 10 cos(x / 5 - 16 + phi) = 10 cos(x / 5). Thus, phi = 16.

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

A wave pulse is propagating with speed $c$ towards positive $x-$axis. The shape of pulse at $t=0$, is $y=ae^{-x/b}$ where $a$ and $b$ are constant. The equation of wave is :

  1. $ae^{-\left(\dfrac{x-ct}{b}\right)}$
  2. $ae^{\dfrac{ct+x}{b}}$
  3. $ae^{x-ct}$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A wave pulse moving in the positive x-direction with speed c has the form f(x - ct). Substituting (x - ct) into the initial shape f(x) = a * e^(-x/b) gives a * e^(-(x - ct) / b).

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

A travelling wave is propagating along negative $x-$axis through a stretched string. The displacement of a particle of the string at $x=0$ is $y=a\cos \omega t$. The speed of wave is $c$. The wave equation is :

  1. $y=a\cos \omega t$
  2. $y=2a\cos \omega t$
  3. $y=a\cos \omega$ $\left(t-\dfrac{x}{c}\right)$
  4. $y=a\cos \left(\omega t+\dfrac{\omega x}{c}\right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a wave traveling in the negative x-direction, the argument is (omega * t + k * x). Given k = omega / c, the equation is y = a * cos(omega * t + omega * x / c).

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

Transverse waves on a string have wave speed $8.00$ m/s, amplitude $0.0700\  m$ and wavelength $0.32\  m$. The waves travel in the negative x-direction and $t = 0$ the $x = 0$ end of the string has its maximum upward displacement. Write a wave function describing the wave.

  1. $\displaystyle \,y\,(x,\,t)\,=\,(0.07\,m)\,sin\,2\,\pi\,\left ( \frac{x}{0.32\,m}\,+\,\frac{t}{0.04\,s} \right )$
  2. $\displaystyle \,y\,(x,\,t)\,=\,(77\,m)\,cos\,2\,\pi\,\left ( \frac{x}{0.32\,m}\,+\,\frac{t}{0.04\,s} \right )$
  3. $\displaystyle \,y\,(x,\,t)\,=\,(0.7\,m)\,sin\,4\,\pi\,\left ( \frac{x}{0.32\,m}\,+\,\frac{t}{0.04\,s} \right )$
  4. $\displaystyle \,y\,(x,\,t)\,=\,(0.97\,m)\,sin\,2\,\pi\,\left ( \frac{x}{0.32\,m}\,+\,\frac{t}{0.04\,s} \right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A left travelling transverse wave given by $y=Asin(kx+\omega t)$

where wave number,$k=\dfrac{2\pi}{\lambda}=\dfrac{2\pi}{0.32}rad/m$
Speed of wave=$\lambda\nu=8m/s$
$\implies \nu=\dfrac{8}{0.32}Hz=25Hz$
$\implies \omega=2\pi\nu=\dfrac{2\pi}{0.04s}$
Thus $y=(0.07m)sin2\pi(\dfrac{x}{0.32m}+\dfrac{t}{0.04s})$

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

Transverse waves on a string have wave speed $12.0$ m/s, amplitude $0.05\  m$ and wavelength $0.4\  m$. The waves travel in the $+ x$ direction and at $t = 0$, the $x = 0$ end of the string has zero displacement and is moving upwards. Find the transverse displacement of a point at x = 0.25 m at time t = 0.15 s.

  1. $-4.54 \ cm$
  2. $-5.54 \ cm$
  3. $-3.54 \ cm$
  4. $-9.54 \ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A wave traveling in +x direction is represented by $y=Asin(\omega t-kx)$

where $\omega=2\pi \nu$
and $k=\dfrac{2\pi}{\lambda}$
We know that speed of wave=$v=\lambda\nu$
Thus here
$\omega=2\pi\times \dfrac{v}{\lambda}=2\pi\times \dfrac{12}{0.4}=60\pi s^{-1}$
and $k=\dfrac{2\pi}{0.4}=5\pi m^{-1}$
Thus wave is $y=(0.05m)sin((60\pi s^{-1})t-(5\pi m^{-1})x)$
Thus the displacement of point at $x=0.25m$ and $t=0.15s$ can be found by putting the values in the equation of wave.
Thus $y(x=0.25m,t=0.15s)=-0.0354m=-3.54cm$

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

Three component sinusoidal waves progressing in the same direction along the same path have the same period, but their amplitudes are $A$, $\displaystyle \frac{A}{2}$ and $\displaystyle \frac{A}{3}$ respectively. The phase of the variation at any position $x$ on their path at time $t = 0$ are $0$, $\displaystyle -\frac{\pi}{2}$ and $-\pi$ respectively. Find the amplitude and phase of the resultant wave.

  1. $\displaystyle \frac{5}{6} A$, $\displaystyle -tan^{-1} \left (\frac{3}{4} \right )$
  2. $\displaystyle \frac{7}{6} A$, $\displaystyle -tan^{-1} \left (\frac{3}{4} \right )$
  3. $\displaystyle \frac{5}{6} A$, $\displaystyle -tan^{-1} \left (\frac{1}{4} \right )$
  4. $\displaystyle \frac{7}{6} A$, $\displaystyle -tan^{-1} \left (\frac{1}{4} \right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The waves with opposite phases superimpose to give a resultant wave of amplitude, $A-\dfrac{A}{3}=\dfrac{2A}{3}$ with phase angle $0$ at time $t=0$.

This superimposes with wave of amplitude $\dfrac{A}{2}$ in with phase $-\dfrac{\pi}{2}$ at $t=0$.

Hence, the resulting wave has amplitude $\sqrt{(\dfrac{2A}{3})^2+(\dfrac{A}{2})^2}=\dfrac{5}{6}A$
The phase of the resulting wave is $tan^{-1}\dfrac{-\dfrac{A}{2}}{\dfrac{2A}{3}}$$=-tan^{-1}\dfrac{3}{4}$

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

Two wave pulses travel in opposite directions on  a string and approach each other. The shape of one pulse is inverted with respect to the other.

  1. The pulse will collide with each other and vanish

    after collision.

  2. The pulses will reflect each other, that is pulse

    going towards right will finally move towards left

    and vice versa.

  3. The pulses will pass through each other but their

    shapes will be modified.

  4. The pulses will pass through each other without

    any change.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When two wave pulses travel on a string and meet, they obey the principle of superposition. They pass through each other without being permanently altered, maintaining their original shapes and velocities after the interaction.

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

 Standing waves are generated on string laded with a cylindrical body. If the cylinder immersed in water, the length of the loops changes by a factor of 2.2. The specific gravity of the material of the cylinder is 

  1. 1.11

  2. 2.15

  3. 2.50

  4. 1.26
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The frequency of a vibrating string is proportional to the square root of tension. When immersed in water, the tension changes due to the buoyant force. The ratio of frequencies (or loop lengths) relates to the density of the object and the fluid, leading to the specific gravity calculation.

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

In a string the speed of wave is 10 m/s and its frequency is 100 Hz . The value of the phase difference at a distance 2.5 cm will be :

  1. ${ \pi }/{ 2 }$
  2. ${ \pi }/{ 8 }$
  3. ${ 3\pi }/{ 2 }$
  4. ${ 2\pi }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Speed of wave $(v) = 10 m/s$

Frequecy$(\gamma)=100Hz$
Wavelength$(\lambda)=\dfrac{10}{100}=\dfrac{1}{10}ms$
$2\pi$ phase is covered in $\dfrac{1}{10}m$
Hence, at distance of 2.5 m, the phase is $\dfrac{2\pi \times 0.025}{0.1}=\dfrac{\pi}{2}$



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

A travelling wave travelled in string in +x direction with 2 cm/s, particle at x=0 oscillates according to equation y (in mm) $= 2\sin { \left( \pi t+{ \pi  }/{ 3 } \right)  }$. What will be the slope of the wave at x=3 cm and t=1 s

  1. $-\sqrt { 3 } { \pi }/{ 2 }$
  2. $\tan ^{ -1 }{ \left( -\sqrt { 3 } { \pi }/{ 2 } \right) }$
  3. $-\sqrt { 3 } { \pi }/{ 20 }$
  4. $-\sqrt { 3 } { \pi }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The wave equation is y = 2 sin(pi * t - pi * x / v + pi / 3) where v = 2 cm/s. The slope is the partial derivative dy/dx. Calculating dy/dx at x=3 and t=1 yields the negative value of the derivative component.

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

The wave-function for a certain standing wave on a string fixed at born ends is y(x, t) = 0.5 sin (0.025$\pi$x) cos 500 t where x and y are in centimeters and t is in seconds The shortest possible length of the string is

  1. 126 cm

  2. 160 cm

  3. 40 cm

  4. 80 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\phi \vec { B } .\vec { dl } ={ \mu  } _{ 0 }{ I } _{ enclosed }$.
Inside the hollow pipe, ${ I } _{ enclosed }=0$.
$\therefore$   $\phi \vec { B } .\vec { dl } =0$
$\Rightarrow$  $B=0$ inside the pipe $\longrightarrow \left( A \right) $.
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A uniform wire of length 20 m and weighing 5 kg hangs vertically. If g=10 $ms^{-2}$, then the speed of transverse waves in the middle of the wire is

  1. $10 ms ^{-1}$
  2. $10\sqrt2 ms ^{-1}$
  3. $15ms ^{-1}$
  4. $2 ms ^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,

$m=5kg$
$l=20m$
$\mu=\dfrac{m}{l}=\dfrac{5}{20}=0.25kg/m$
$g=10m/s^2$
Tension in the middle of the wire, $T=\dfrac{m}{2}g$
$T=\dfrac{5}{2}\times 10=25N$
Velocity, $v=\sqrt{\dfrac{T}{\mu}}$
$T=\sqrt{\dfrac{25}{0.25}}=10m/s$
The correct option is A.