Tag: energy in wave motion

Questions Related to energy in wave motion

If the frequency and amplitude of a transverse wave on a string are both doubled, then the amount of energy transmitted through the string is

energy in wave motion oscillation and waves waves physics

  1. doubled

  2. become 4 time

  3. becomes 16 times

  4. becomes 32 times

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Correct Option: C
Explanation:

Energy transmitted through the string $E  \propto  \nu^2  A^2$

Now, $\nu' = 2  \nu                      A' = 2  A$
Thus, $\dfrac{E'}{E}  = \dfrac{(\nu')^2  (A')^2}{\nu^2    A^2} = \dfrac{4  \nu^2  \times  4  A^2}{\nu^2   A^2}$

$\implies    E' = 16    E$

A string of per unit length $\mu$ is clamped at both ends such that one end of the string is at $x = 0$ and the other is at $x = \ell$. When string vibrates in fundamental mode amplitude of the mid-point of the string is a and tension in the tension in the string is $T$. If the total oscillation energy stored in the string is $\displaystyle \,\frac{\pi^2\,a^2\,T}{xl}$. Then the value of $x$ is 

energy in wave motion oscillation and waves waves physics

  1. 1

  2. 2

  3. 3

  4. 4

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Correct Option: D

$y _1 = 88\, sin(\omega t - kx)$ and $y _2 = 6 sin(\omega t + kx)$ are two waves travelling in a string of area of cross-section $s$ and density $\rho$. These two waves are superimposed to produce a standing wave. Find the total amount of energy crossing through a node per second.

energy in wave motion oscillation and waves waves physics

  1. $\displaystyle \frac{2\rho\omega^{3}s}{k}$

  2. $\displaystyle \frac{3\rho\omega^{3}s}{k}$

  3. $\displaystyle \frac{5\rho\omega^{3}s}{k}$

  4. $\displaystyle \frac{6\rho\omega^{3}s}{k}$

Show answer
Correct Option: A
Explanation:

 Here, we will take ,

               $y=y _1 + y _2$
                     = $8sin(\omega t-kx) +6 sin(\omega t+kx)$..........(1)
Now, add and subtract $6sin(\omega t -kx)$ in (1).
                      = $8sin(\omega t-kx) +6sin(\omega t+kx) +6sin(\omega t-kx) -6 sin(\omega t
-kx)$
                      =$2sin(\omega t-kx) +12 sin\omega t cos kx$............(2)
We obtained (2) by solving using trigonometric relations.
energy crossing through node per second = power
                       $P=\dfrac{1}{2}\rho \omega^2 (2)^2 Sv$..................(3)
Now, put $v=\dfrac{\omega}{k}$ in eqn. (3)

  We get , $P= \dfrac{2\rho\omega^3 s}{k}$


Choose the correct alternative(s) regarding standing waves in a string

energy in wave motion oscillation and waves waves physics

  1. particles near the antinode have lesser potential energy than the particles near the node when they reaches at its extreme position

  2. All the particles crosses their mean position simultaneously

  3. Energy and momentum can transmitted through node

  4. Particles near the antinode have lesser kinetic energy than the particles near the node when they crosses their mean position

Show answer
Correct Option: A,B
Explanation:

At the antinode, the tension and hence the elongation in the string in minimum and hence minimum potential energy. While at the nodes, tension is maximum and hence maximum potential energy. All particles perform SHM with same time period and hence since the phase differecnce between any two particles is constant, they cross mean position simultaneously. 

As note is always at rest, energy and momentum cannot be transferred through it.
Particles at antinode have maximum velocity and hence maximum kinetic energy while crossing the mean position.

With the propagation of a longitudinal wave through a material medium, the quantities transmitted in the direction of propagation are

energy in wave motion oscillation and waves waves physics

  1. Energy, momentum and mass

  2. Mass and momentum

  3. Energy and mass

  4. Energy and momentum

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Correct Option: D
Explanation:

Whenever any wave travels through a material medium, the particles of the medium start vibrating about their mean positions.

Every vibrating particle transfers its vibration to its immediate next particle.
In any vibration there exists Kinetic Energy and Potential Energy. As the vibration gets transfered, the energy also gets transfered.
Momentum is defined as the product of mass and velocity. Since the particles have mass and they are in motion, they have momentum. So obviously when vibration is transfered, momentum is also transfered.
In any wave motion, the vibration travels in the forward direction, but no particle actually travels forward. Hence mass doesn't get transfered.
We get an illusion that something is moving forward, although nothing moves forward.

Please note that the question is about a longitudinal wave, then also these same facts are applicable to a transverse wave also.

The amplitude of two waves are in ratio 5 : 2. If all other conditions for the two waves are same, then what is the ratio of their energy densities?

energy in wave motion oscillation and waves waves physics

  1. 5 : 2

  2. 5 : 4

  3. 4 : 5

  4. 25 : 4

Show answer
Correct Option: D
Explanation:

Energy density of wave is given by
$u=2\pi^2n^2pA^2$
or $u \propto A^2$   (As n and p are constant)
$\therefore \dfrac{u _1}{u _2}=\dfrac{A _1^2}{A _2^2}=\dfrac{5^2}{2^2}$
So, $u _1:u _2=25:4$

A progressive wave on a string having linear mass density $\rho$ is represented by $y = A\sin \left (\dfrac {2\pi}{\lambda} x - \omega t\right )$ where $y$ is in $10\ mm$. Find the total kinetic energy (in $\mu l)$ passing through origin from $t = 0$ to $t = \dfrac {\pi}{2\omega}$.
[Take : $\rho = 3\times 10^{-2} kg/ m; A = 1mm; \omega = 100\ rad/ sec; \lambda = 16\ cm]$

energy in wave motion oscillation and waves waves physics

  1. $6$

  2. $7$

  3. $8$

  4. $9$

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Correct Option: D

A clamped string is oscillating in nth harmonic, then 

energy in wave motion oscillation and waves waves physics

  1. total energy of oscillations will be $n^{2}$ times that of fundamental frequency

  2. total energy of oscillations will be $(n-1)^{2}$ times that of fundamental frequency

  3. average kinetic energy of the string over a complete oscillations is half of the total energy of the string

  4. none of these

Show answer
Correct Option: A,C
Explanation:

For a sine wave, $y = A \sin(kx -  \Omega  t)$
Velocity equation for this wave is $V _y =  \Omega A \cos(kx - \Omega  t)$
Kinetic energy = $ d(KE) = 1/2(V _y^2 \times dm) = 1/2(V _y^2 \times  \mu  dx)$, $ \mu $ is the linear mass density.


=> $1/2( \mu  \times  \Omega ^2 \times A^2 \times cos^2(kx -  \Omega  t)) dx$
integrating at $t = 0,$ with limits as $0$ and $ \lambda $, we have

$K.E = 1/4( \mu \times \Omega^2 \times A^2 \times  \lambda $)
Potential energy, $dU = 1/2 (  \Omega  ^2 \times y^2 \times  \mu ) dx $

integrating at $t = 0$, with limits as $0$ and $ \lambda $, we have
$U = 1/4( \mu \times \Omega  ^2 \times A^2 \times  \lambda $)

Total energy $E = K.E + U
$
=> $E = 1/2( \mu  A^2  \lambda $)
Therefore, for the first and fundamental frequency, energy is 
$E _1 = (1/2( \mu  A^2  \lambda ))/n^2$
And clearly from the above derivation, we have, K.E is half the total energy.

To determine the position of a point like object precisely ______ light should be used.

energy in wave motion oscillation and waves waves physics

  1. polarized

  2. short wavelength

  3. long wavelength

  4. intense

Show answer
Correct Option: B
Explanation:
To determine the position of a point like object precisely light of short wavelength should be used.