Tag: wave motion

Quizzes (25) Questions (122)

Questions Related to wave motion

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

The equation of a progressive wave for a wire is: 
$Y=4\sin{\left[\cfrac{\pi}{2}\left(8t-\cfrac{x}{8}\right)\right]}$. If $x$ and $y$ are measured in cm then velocity of wave is :

  1. $64 cm/s$ along $-x$ direction
  2. $32 cm/s$ along $-x$ direction
  3. $32 cm/s$ along $+x$ direction
  4. $64 cm/s$ along $+x$ direction
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\begin{array}{l} w=4\pi  \ K=\dfrac { \pi  }{ { 16 } }  \ v=\dfrac { w }{ K } =64\, m/s\, along\, \, +x-axis \ Hence, \ option\, \, D\, \, is\, correct\, \, answer. \end{array}$

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

An open tube is in resonance with string (frequency of vibration of tube in $n _{0}$. If tube is dipped on water is that 75% of length of tube is inside water, then the ratio of the frequency of tube to string now will be 

  1. 1

  2. 2
  3. $\dfrac{2}{3}$
  4. $\dfrac{3}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
For open tube no $ = \dfrac{V}{2l} $
For closed tube length available for resonance
$ l^{1} ,l\times \dfrac{25}{100} = \frac{l}{4} $
fundamental frequency of water filled tube 
$ n _{1}\dfrac{V}{4l^{1}} = \frac{V}{4(l/4)} $ $(\because l^{1}= l/4) $
$ \therefore \dfrac{V}{l} = 2n _{0} = 1 $
$ = \dfrac{n}{n _{0}} = 2 $
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

The equation of a standing wave in a string fixed at both ends is given as $ y =  A \quad sin \quad  kx \quad cos \quad \omega t $
The amplitude and frequency of a particle vibrating at the mid of an antiode and a node are respectively

  1. $A,\dfrac{\omega }{{2\pi }}$
  2. $\dfrac{A}{{\sqrt 2 }},\dfrac{\omega }{{2\pi }}$
  3. $A,\dfrac{\omega }{{\pi }}$
  4. $\sqrt 2 A,\dfrac{\omega }{{2\pi }}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a standing wave y = A sin(kx) cos(omega t), the amplitude of a particle at position x is A_particle = |A sin(kx)|. A point midway between a node and an antinode corresponds to kx = pi/4. At this point, the amplitude is A * sin(pi/4) = A / sqrt(2). The frequency of vibration of any particle in a standing wave is the same as the wave frequency, which is omega / (2 pi).

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

A wire of length l , area of cross section A  and young's modules of elasticity  y is  suspended from the roof of a building. A  block of mass m is attached at lower end of the wire. if the block is displaced from its mean position and then released the block starts  oscillating. Time period of these oscillation will be

  1. $2\pi \sqrt { \frac { Al }{ mY } } $
  2. $2\pi \sqrt { \frac { AY }{ ml } } $
  3. $2\pi \sqrt { \frac { ml }{ YA } } $
  4. $2\pi \sqrt { \frac { m }{ YAl } } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When a block of mass m is suspended from an elastic wire of length l, area A, and Young's modulus Y, the effective spring constant of the wire is k = YA / l. The time period of vertical oscillations of the mass-spring system is given by T = 2 pi sqrt(m / k) = 2 pi sqrt(ml / YA).

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

Which of the following equations represents a transverse wave travelling along -y axis?

  1. $x = A\sin\ (\omega t\ -\ ky)$
  2. $x= A\ sin\ (\omega t\ +\ ky)$
  3. ${ y } _{ 0 }\ =A\sin\ (\omega t - kX )$
  4. ${ y } _{ 0 } = A\ sin (\omega t + kX )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} For\, \, negative\, \, y-axis \ sign\, \, of\, \, \omega t\, \, & \, \, ky\, \, should\, \, be\, \, same\,  \ x=A\sin  \left( { \omega t+ky } \right)  \ Hence, \ option\, \, B\, \, is\, correct\, \, naswer. \end{array}$

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

The displacement from the position of equilibrium of a point $4\ cm$ from a source of sinusoidal oscillations is half the amplitude at the moment $t=\dfrac{T}{6} (T$ is the time period$)$. Assume that the source was at mean position at $t=0$. The wavelength of the running wave is 

  1. $0.96\ m$
  2. $0.48\ m$
  3. $0.24\ m$
  4. $0.12\ m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Going by the data given to us, this wave is sinusoidal in nature and the wave equation takes the form of
$y = A\sin( \omega t - kx),$ as it is given that at $t = 0,$  the source is at mean position.
Here$,\ x = 4\ cm = 0.04\ m$
$y = A/2$
Amplitude $= A$
$t = \dfrac{T}{6}$
We know that $ \omega  = 2 \dfrac{ \pi }{T}$
$\Rightarrow \dfrac{A}{2} = A\sin((2  \pi  / T)(T/6) - 0.04k)$
$\sin((2 \pi  / T)(T/6) - 0.04k) = 1/2$
$\Rightarrow ((2  \pi / T)(T/6) - 0.04k) =  \pi / 6$
$k =  \pi  / 0.24$
wavelength $ \lambda = 2\pi  / k = 0.48\ m$

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

A string of length 1 m fixed at one end and on the other end a block of mass M=4 kg is suspended.The string is set into vibrations and represented by equation, Y=$6\sin \left( {\dfrac{{\pi x}}{{10}}} \right)\;\cos \;100\;\pi t,$  where x and y are in cm an in seconds.
Find the number of loops formed in the string.

  1. 3

  2. 4

  3. 5

  4. 6
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

A travelling wave on a light on a tight string is described by the equation $y=A\sin (kx-\omega t)$. if tension in the string is $F$ then total energy stored in the string having from $x=0$ to $x=2\pi/k$ is 

  1. $\pi FA^{2}$
  2. $\pi kFA^{2}$
  3. $\pi k^{2}FA$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The total energy stored in one wavelength of a travelling wave on a string is given by E = (1/2) * mu * omega^2 * A^2 * lambda, where mu = F / v^2 and v = omega / k. Substituting these relations yields E = pi * k * F * A^2.