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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

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}$

Multiple choice physics energy production scattering of light and its applications some natural phenomena of light effects of light

A beam of monochromatic light first travels through glass of R.I. $1.5$ and then through water (R.I. $=\dfrac{4}{3}$). If the difference in their wavelengths in the two media is $40$nm, then the wavelength of the monochromatic light in vacuum will be?

  1. $4000\overset{o}{A}$
  2. $4400\overset{o}{A}$
  3. $4800\overset{o}{A}$
  4. $5200\overset{o}{A}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics human eye and colourful world scattering of light and its applications some natural phenomena of light effects of light

One can not see through fog because:

  1. fog absorbed light

  2. light is scattered by the droplets in fog

  3. light surfers total reflection by the droplets in the fog

  4. the refractive index of fog is in infinity
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We can not see clearly through fog because the light is scattered by the droplets in the fog. The light is scattered by very small particles, this phenomenon is called Tyndall effect.

Hence the option B is the right answer

Multiple choice physics human eye and colourful world scattering of light and its applications some natural phenomena of light effects of light

The rising and setting of sun appear red because of : 

  1. refraction

  2. reflection

  3. defraction

  4. scattering
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The reddish appearance of the sun at sunrise or sunset is due to scattering of light by the molecules of air and other fine particles in the atmosphere.


Hence the option D is the right answer