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

A $100$ Hz sinusoidal wave is travelling in the positive x-direction along a string with a linear mass density of $3.5 \times 10^{-3}$ kg/m and a tension of $35$ N. At time t = 0, the point x = 0 has zero displacements and the slope of the string is $\pi/20$. Then select the wrong alternative

  1. Velocity of wave is $100$ m/s
  2. Angular frequency is $(200 \pi)$ rad /s
  3. Amplitude of wave is $0.025$ m
  4. Propagation constant is $(4 \pi)$ $m^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

The wave function for the wave pulse is $ Y (X,t) = \frac {0.1a^3}{a^2 +(X-Vt)^2}  with a = 4 cm. At X = 0 $ The displacement y (x,t) is observed to decreases from its maximum value to half of that value in time $ t = 2 \times 10^{-3} s $ choose the correct statement 

  1. The wave pulse is moving is negative X direction with speed 10 m/s

  2. The wave pulse is moving is positive X direction with speed 10 m/s

  3. The wave pulse is moving is negative X direction with speed 20 m/s

  4. The wave pulse is moving is positive X direction with speed 20 m/s

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

The  equation of a transverse wave travel on a rope is given   y = 10 sin $\pi$(0.01x - 2.00t) where y and x in cm and t in seconds.The maximum transverse  speed  of a particle in the rope about 

  1. 62.8 cm / s

  2. 75 cm / s

  3. 100 cm / s

  4. 121 cm / s

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

The wave equation is y = 10 sin(0.01 * pi * x - 2 * pi * t). The transverse velocity is v_y = dy/dt = -10 * 2 * pi * cos(0.01 * pi * x - 2 * pi * t). The maximum transverse speed is 20 * pi = 20 * 3.14 = 62.8 cm/s.

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

A wave represented by equation $y = 2(mm) \, sin \, [4 \pi (sec^{-1}) t - 2 \pi (m^{-1}) X]$ is superimposed with another wave $y = 2 (mm) sin [4 \pi (sec^{-1}) t + 2 \pi (m^{-1}) x + \pi/3]$ on a tight string.
Phase difference between two particles with are located at $x _1 = 1/7$ and $x _2 = 5/12$ is :

  1. $0$
  2. $\dfrac{5 \pi}{6}$
  3. $\pi$
  4. $\dfrac{5 \pi}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

${y _1} = 2\sin \left[ {4\pi t - 2\pi x} \right]$

${y _2} = 2\sin \left[ {4\pi t - 2\pi x} \right]$
$y = {y _1} + {y _2} = 2\left[ {2\sin 4\pi t\,\,\cos 4\pi x} \right]$
$ = 4\sin 4\pi t\cos 4\pi x$
$y = 4\cos 4\pi \sin 4\pi t$
Amp pass$\left| {4\cos 4\pi x} \right| = 4$
$ \Rightarrow \cos 4\pi x =  \pm 1$
$ = 0,\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{4}{4},\frac{5}{4},\frac{6}{4}.......$
$\therefore for\,{x _1} = \frac{1}{9}and\,{x _2} = \frac{5}{{12}}$
hence,
phase difference is$\pi$

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

A travelling wave on a string is given by $y = A$ $A \sin \left[ \alpha x + \beta t + \cfrac { \pi } { 6 } \right]$ The displacement and velocity of oscillation of a point $\alpha =$ $0.56 / \mathrm { cm } , \beta = 12 / \mathrm { sec }$ $A = 7.5 \mathrm { cm } , x = 1$ $\mathrm { cm }$ and $\mathrm { t } = 1 \mathrm { s }$ is

  1. $4.6 \mathrm { cm } , 46.5 \mathrm { cm } s ^ { - 1 }$
  2. $3.75 \mathrm { cm } , 77.94 \mathrm { cm } \mathrm { s } ^ { - 1 }$
  3. $1.76 \mathrm { cm } , 7.5 \mathrm { cms } ^ { - 1 }$
  4. $7.5 \mathrm { cm } , 75 \mathrm { cm } \mathrm { s } ^ { - 1 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} y=7.5\sin  \left[ { 0.56\alpha +12t+\dfrac { \pi  }{ 6 }  } \right]  \ at\, \, x=1 \ & \, \, t=1 \ y=7.5\sin  \left[ { 0.56+12+\dfrac { \pi  }{ 6 }  } \right]  \ =7.5\sin  \left[ { 12.56+\dfrac { \pi  }{ 6 }  } \right]  \ =7.5\sin  \left[ { 4\pi +\dfrac { \pi  }{ 6 }  } \right]  \ =3.75\, \, cm \end{array}$

$\therefore$ Option $B$ is correct .

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

A sine wave is travelling in a medium. The minimum distance between the two particles. always having same speed is 

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

In a sine wave particles that are separated by a distance of odd multiple of half

the wave length move with same speed and but in opposite direction. 
The minimum separation is $\frac{\lambda }{2}$
Option C.

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

If a string is stretched by $\dfrac{L}{20}$ then velocity of wave is $V$. When string is stretched by $\dfrac{L}{10}$ then velocity becomes

  1. $\dfrac{V}{\sqrt 2}$
  2. $V$
  3. $2V$
  4. $\sqrt 2 V$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a stretched string, T = Y * A * (delta_L / L). Wave speed v = sqrt(T / mu) = sqrt(Y * A * delta_L / (L * mu)). Since v is proportional to sqrt(delta_L), doubling the extension delta_L (from L/20 to L/10) increases the velocity by a factor of sqrt(2).

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

Sinusoidal waves 5.00 cm in amplitude are to be transmitted along a string having a linear mass density equal to 4.00 * $10^-2 kg/m$. If the source can deliver a average power of 90 W and the string is under a tension of 100 N,then the highest frequency at which the source can operate is (take $\pi^2 = 10)$:

  1. 45 Hz

  2. 50 Hz

  3. 30 Hz

  4. 62 Hz

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Average power P = 2 * pi^2 * f^2 * A^2 * mu * v. Tension T = 100 N, mu = 0.04 kg/m, so v = sqrt(100 / 0.04) = 50 m/s. P = 90 W, A = 0.05 m. 90 = 2 * 10 * f^2 * (0.05)^2 * 0.04 * 50. 90 = 20 * f^2 * 0.0025 * 2 = 0.1 * f^2. f^2 = 900, so f = 30 Hz.

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

The equation of a stationary wave in a string is y =(4) sin[($3.14m^-1$)x] cos ${\omega}t$. (mm)
Select the correct alternative(s).

  1. the amplitude of component waves is 2 mm

  2. the amplitude of component waves is 4 mm

  3. the smallest possible length of string is 0.5 m

  4. the smallest possible length of string is 1.0 m

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation of stationary wave in a string is the amplitude of component waves is $4$mm.

Hence, option $B$ is correct ansnwer.

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

A wave represented by a given equation $y (x,t) = a \sin (\omega t - kx)$superimposes on another wave giving a stationary wave having antinode at $x = 0 $ then the equation of the another wave is 

  1. $y = - a \sin (\omega t - kx)$
  2. $y = a \sin (\omega t + kx)$
  3. $y = - a \sin (\omega t + kx)$
  4. $y = - a \cos (\omega t + kx)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} a\sin  \left( { wt-kx } \right) +{ y _{ 1 } }=2a\sin  \cot  \cos  kx \ \Rightarrow y=a\sin  \left( { wt+kx } \right)  \ Ans.\, \, (B) \end{array}$

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

A travelling wave passes  point of observation. At this point, the time interval between successive crests is $0.2\, s$ and 

  1. The wavelength is $5\, m$
  2. The frequency is $5\, Hz$
  3. The velocity of propagation is $5\, m/s$
  4. The wavelength is $0.2\, m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} Accorrding\, \, to\, \, question....................... \ Here, \ Difference\, between\, two\, successive\, crest\, is\, 2p. \ passes\, difference\, (\Delta \varphi )=\frac { { 2\pi  } }{ T }  \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, T=time\, { { int } }erval(\Delta t) \ \therefore \, \, \, n=2\pi =\frac { { 2\pi  } }{ T } \times 0.2 \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Rightarrow \frac { 1 }{ T } =5{ \sec ^{ -1 }  } \ \Rightarrow n=5Hz \ So\, the\, correct\, option\, is\, B. \end{array}$

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

The equation of standing wave in a stretched string is given by by y = 5sin($\frac{{\pi}{x}} {3}$) cos $(40{\pi}t)$, where x and y are in cm and t in second. The separation between two consecutive nodes is (in cm) 

  1. 1.5

  2. 3

  3. 6

  4. 4

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} y=5\sin  \left( { \frac { { \pi x } }{ 3 }  } \right) .\cos  \left( { 4\pi t } \right)  \ Here\, in\, above\, equation\, \, k=\frac { \pi  }{ 3 }  \ \frac { { 2\pi  } }{ \lambda  } =\frac { \pi  }{ 3 }  \ \lambda =6cm \ Hence,\, dis\tan  ce\, between\, 2\, nodes\, is\, \frac { \lambda  }{ 2 } =3cm \end{array}$

Hence, the option $B$ is the correct answer.