Tag: problems on properties of waves

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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

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. $1/6$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

y1 = 10 sin(3pi*t + pi/4), amplitude = 10. y2 = 5(3 sin(3pi*t) + sqrt(3) cos(3pi*t)). Using R = sqrt(A^2 + B^2 + 2AB cos(phi)), y2 = 5 * sqrt(3^2 + sqrt(3)^2) * sin(...) = 5 * sqrt(9+3) = 5 * sqrt(12) = 10 * sqrt(3). Ratio = 10 / (10 * sqrt(3)) = 1/sqrt(3).

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

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

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

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. $4{I} _{0}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Resultant intensity for two interfering waves of intensity I_0 and phase difference phi is I = I_1 + I_2 + 2 * sqrt(I_1 * I_2) * cos(phi). With I_1 = I_2 = I_0 and phi = 60 degrees, cos(60) = 1/2, yielding I = I_0 + I_0 + I_0 = 3 I_0.

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

The displacement of an elastic wave is given by the function $y= 3\ sin \omega t+4\ cos\omega t$, where $y$ is in $cm$ and $t$ is in $s$. The resultant amplitude is 

  1. $3 cm$
  2. $ 4 cm$
  3. $ 5 cm$
  4. $7 cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here, we have given: $y=3sin\omega t+4cos\omega t$


So, two components are there, $3sin\omega t $ and $4cos\omega t$


where, individual amplitudes are given by
$A _1= 3 cms$ and $A _2=4 cms .$

so , resultant amplitude will be, 
$A=\sqrt{A _1^2 +A _2^2}=\sqrt{3^2+4^2}$

$A=5 cms$


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}$