Tag: waves

Questions Related to waves

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 $


Correct Option: B

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


Correct Option: A

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. $none\ of\ these$


Correct Option: B

A wave has a wavelength of 3m. The distance between a crest and adjacent trough is

  1. 0.75 m

  2. 1.5 m

  3. 3 m

  4. 1 m


Correct Option: C

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. $2 \pi$


Correct Option: C
Explanation:

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

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

Two simple harmonic motions are represented by the equations 
$y _1=10\sin \left(3\pi t+\dfrac{\pi}{4}\right)$
and $y _2=5(3\sin 3\pi t+\sqrt 3 \cos 3\pi t)$ Their amplitudes are in the ratio of :

  1. $\sqrt 3$

  2. $1/\sqrt 3$

  3. $2$

  4. $1/6$


Correct Option: B

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 .$


Correct Option: A
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 $\Delta x=3\lambda/4$
$\Delta \phi=2\pi/\lambda \times 3\lambda/4=1.5\pi\, rad$.

Vibrations of period 0.25 s propagate along a straight line at a velocity of 48 cm/s. One second after the emergence of vibrations at the initial point, displacement of the point, 47 cm from it is found to be 3 cm. Then,

  1. amplitude of vibrations is 6 cm.

  2. amplitude of vibrations is $3 \sqrt{2} cm.$

  3. amplitude of vibrations is 3 cm.

  4. None of the above


Correct Option: A
Explanation:

The wavelength of the wave can be found by using $\dfrac{\lambda}{T}=v$

$\implies \lambda=vT=48cm/s\times 0.25s=12cm$
Four full wavelengths complete at a distance of 48cm.
Thus a point 47cm lag by a phase difference of $\dfrac{2\pi}{\lambda}(48cm-47cm)=\dfrac{\pi}{6}$
Let the amplitude of vibrations be $A$.
Thus the displacement at the given point=$Asin(\dfrac{\pi}{6})=\dfrac{A}{2}=3cm$
$\implies A=6cm$
Thus correct answer is option A.

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 $\left [\pi (6t-\frac{X}{60}) \right ]$


Correct Option: C

If two waves, each of intensity ${I} _{0}$, having the same frequency but differing by a constant phase angle of ${60}^{o}$, superpose at a certain point in space, then the intensity of resultant wave is:

  1. $2{I} _{0}$

  2. $\sqrt{3}{I} _{0}$

  3. $3{I} _{0}$

  4. $4{I} _{0}$


Correct Option: B