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 problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

Which of the following equations does not represent a progressive wave ?

  1. $y=Asin[\omega (t-\frac { x }{ v } )]$
  2. $y=Asin[ \frac { 2pi }{ \lambda }(vt-x)] $
  3. $y=Asin[2\pi (\frac {t}{T}-\frac { x }{ \lambda } )]$
  4. $y=Asin[2\pi (\frac {t}{T}-\frac { x }{ v } )]$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A progressive wave must be a function of (vt - x) or (t - x/v). Option D uses (t/T - x/v), which is dimensionally inconsistent if T is period and v is velocity, as t/T is dimensionless but x/v is time.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

A traveling wave is represented by the equation $ y = \frac{1}{10} sin(60 t + 2x) $, where x and y in meters and t is in second . this represents a wave
(1) of frequency $ \frac {30}{\pi} Hz $
(2) of wavelength $ \pi m $
(3)of amplitude 10 cm
(4) moving in the positive x direction
pick out the correct statements from the above.

  1. 1, 2, 4

  2. 1, 3, 4

  3. 1, 2, 3

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

y = 0.1 sin(60t + 2x). omega = 60, k = 2. Frequency f = omega/2pi = 60/2pi = 30/pi Hz. Wavelength lambda = 2pi/k = 2pi/2 = pi m. Amplitude = 0.1 m = 10 cm. The wave moves in negative x direction because of the + sign.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

A wave equation which given the displacement along the Y direction is given by $y = 10^{-4} \sin (60t+2x)$ where x and y are in meters and t is time in seconds. This represents a wave 

  1. Traveling with a velocity of 30 m/s in the negative x direction
  2. Of wavelength $\pi $ metre
  3. Of frequency 30/$\pi $hertz
  4. Of amplitude $10^{ -4 }$ metre
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

For a wave $ y= y _0 sin ( \omega t - kx ) $, for what value of $ \lambda $ , is the maximum particle velocity equal to two times the wave velocity :-

  1. $ \pi y _0 $
  2. $ 2 \pi y _0 $
  3. $ \pi y _0/2 $
  4. $4 \pi y _0 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Particle velocity v_p = dy/dt = y0 * omega * cos(omega*t - kx). Max particle velocity = y0 * omega. Wave velocity v_w = omega/k. Given y0 * omega = 2 * (omega/k), so y0 = 2/k. Since k = 2pi/lambda, y0 = 2 / (2pi/lambda) = lambda/pi. Thus, lambda = pi * y0.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

The equation $y = a \sin^2 \left(2 \pi nt - \dfrac{2\pi x}{\lambda}\right)$ represents a wave with

  1. Amplitude $a$, frequency $n$ and wavelength $\lambda$
  2. Amplitude $a$, frequency $2n$ and wavelength $2\lambda$
  3. Amplitude $a/2$, frequency $2n$ and wavelength $\lambda$
  4. Amplitude $a/2$, frequency $2n$ and wavelength $\lambda/2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using sin^2(theta) = (1 - cos(2*theta))/2, the equation becomes y = a/2 - (a/2)cos(4*pi*n*t - 4*pi*x/lambda). This represents a wave with amplitude a/2, frequency 2n, and wavelength lambda/2 (since k = 4*pi/lambda = 2*pi/lambda_new).

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

The equation $y =A\cos^2\left(2\pi\, nt -2\pi \dfrac{x}{\lambda}\right)$ represents a wave with

  1. amplitude $A/2$, frequency $2n$& wavelength $\lambda/2$
  2. amplitude $A/2$, frequency $2n$& wavelength $\lambda$
  3. amplitude $A$, frequency $2n$& wavelength $2\lambda$
  4. amplitude $A$, frequency $n$& wavelength $\lambda$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using cos^2(theta) = (1 + cos(2*theta))/2, the equation becomes y = A/2 + (A/2)cos(4*pi*n*t - 4*pi*x/lambda). This corresponds to amplitude A/2, frequency 2n, and wavelength lambda/2.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

The equation of a wave is given by 
$ Y\quad =\quad A\quad sin\quad \omega \left( \frac { x }{ v } -k \right)  $
Where $ \omega $ is the angular velocity and v is the linear velocity.The dimensions of K is

  1. LT

  2. T
  3. $ T^{-1} $
  4. $ T^2 $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In the argument omega(x/v - k), the dimensions of omega are T^-1. For the argument to be dimensionless, (x/v - k) must have dimensions of T. Since x/v is distance/velocity = time, k must also have dimensions of time.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

The equation of a progressive wave is $Y= a sin(200 t-x)$, where x is in meter and t is in second. The velocity of wave is

  1. $200 $ m/sec
  2. $100 $ m/sec
  3. $50 $ m/sec

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

The wave equation is y = a sin(200t - x). The standard form is y = a sin(omega*t - k*x). Here omega = 200 and k = 1. Wave velocity v = omega/k = 200/1 = 200 m/s.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

The equation of a wave travelling on a stretched string is :
$y=4\sin 2\pi \left(\dfrac{t}{0.02}-\dfrac{x}{100}\right)$
Here $x$ and $y$ are in $cm$ and $t$ is in second. the relative deformation amplitude of medium is :

  1. $0.02\pi$
  2. $0.08\pi$
  3. $0.06\pi$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The relative deformation (strain) is dy/dx. y = 4 sin(2pi*t/0.02 - 2pi*x/100). dy/dx = 4 * (-2pi/100) * cos(...) = -0.08pi * cos(...). The amplitude of this is 0.08pi.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

A source oscillates with a frequency 25 Hz and the wave propagates with 300 m/s. Two points A and B are located at distances 10 m and 16 m away from the source. The phase difference between A and B is 

  1. $\displaystyle \frac{\pi}{4}$
  2. $\displaystyle \frac{\pi}{2}$
  3. $\pi$
  4. $2 \pi$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Wavelength of the wave=$\lambda=\dfrac{v}{\nu}=\dfrac{300}{25}=12m$

Distance between the two points=$16m-10m=6m=\dfrac{\lambda}{2}$
$=\dfrac{2\pi}{\lambda}\dfrac{\lambda}{2}=\pi$

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

For the travelling harmonic wave  $y(x,t)=2.0 cos $ $ 2\pi $ (10t-0.0080 x+0.35 ) where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of $x$

  1. $x=4 m,\ \ \Delta\phi=6.4π \ rad $
  2. $0.5 m,\ \ \ \ \ \Delta\phi=0.6π \, rad $
  3. $ \displaystyle \lambda /2 ,\ \ \ \ \ \ \ \Delta\phi= .6π \ rad$
  4. $ \displaystyle 3\lambda /4,\ \ \ \ \ \Delta\phi= 2.5π \ rad .$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation for a travelling harmonic wave is given as:

$y(x, t)=2.0\,cos\,2\pi(10t-0.0080x+0.35)$
             $=2.0\,cos(20\pi t-0.016\pi x+0.70\pi)$
Where,
Propagation constant, $k = 0.0160\pi$
Amplitude, $a=2\,cm$
Angular frequency, $\omega =20\pi\,rad/s$
Phase difference is given by the relation:
$\phi =kx=2\pi/\lambda$

(a) For $\Delta x=4m= 400 cm$
$\Delta \phi = 0.016\pi\times 400=6.4\pi\, rad$

(b) For $\Delta x=0.5 m = 50 cm$
$\Delta \phi = 0.016\pi \times 50 = 0.8\pi\, rad$

(c) For $\Delta x=\lambda/2$
$\Delta \phi=2\pi/\lambda \times \lambda/2=\pi\, rad$

(d) For $\Delta x=3\lambda/4$
$\Delta \phi=2\pi/\lambda \times 3\lambda/4=1.5\pi\, rad$.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

A wave travelling in positive X-direction with A = 0.2 m velocity = 360 m/s and $\lambda$= 60 m, then correct expression for the wave is : -

  1. y = 0.2 sin $\left [ 2\pi (6t+\frac{X}{60}) \right ]$
  2. y = 0.2 sin $\left [\pi (6t+\frac{X}{60}) \right ]$
  3. y = 0.2 sin $\left [ 2\pi (6t-\frac{X}{60}) \right ]$
  4. y = 0.2 sin $\left [\pi (6t-\frac{X}{60}) \right ]$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The general equation for a wave moving in the positive x-direction is y = A sin(2*pi*(ft - x/lambda)). Given A = 0.2, f = velocity/lambda = 360/60 = 6 Hz, and lambda = 60, the equation becomes y = 0.2 sin(2*pi*(6t - x/60)).

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

Equations of a stationary wave and a travelling wave are $y _1=1\,sin(kx)\,cos (\omega t)$ and $y _2=a\,sin\,(\omega t-kx)$.The phase difference between two points $x _1=\dfrac{\pi}{3k}$ and $x _2=\dfrac{3 \pi}{2k}$ is $\phi _1$ for the first wave and $\phi _2$ for the second wave.The ratio $\dfrac{\phi _1}{\phi _2}$ is

  1. 1

  2. 5/6

  3. 3/4

  4. 6/7
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Phase difference between two points in a standing wave =$n\pi$

Where n is number of nodes between two points.
Given points are $x _1 = \cfrac{\pi}{3k} = \cfrac{60}{k}$
$x _2 = \cfrac{3\pi}{2k} = \cfrac{210}{k}$
Equation of the standing wave
$ y _1 = a \sin kx \cos \omega t$
At node points $ kx =n\pi$
$ x = \cfrac{n\pi}{k} \quad (n=0,1,2,3...)$
So nodes are =$ \cfrac{\pi}{k} , \cfrac{2\pi}{k} ....$
$ =  \cfrac{180}{k} , \cfrac{360}{k} ....$
Since there is only one node between phase difference  $ \phi _1 = \pi$
For travelling wave $ \phi _2  = \cfrac{2\pi}{\lambda} \triangle x$
From the equation 
$y _2 = a \sin (\omega t - kx)$
$ k = \cfrac{2\pi}{\lambda}$
$ \therefore \phi _2 = k[x _2 - x _1] = k[\cfrac{3\pi}{2k} - \cfrac{\pi}{3k}] = \cfrac{7}{6}\pi$
$ \therefore \cfrac{\phi _1}{\phi _2} = \cfrac{\pi}{\cfrac{7}{6}\pi} = \cfrac{6}{7}$