Tag: waves

Questions Related to waves

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

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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

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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

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

Two open pipes of length $20$ cm and $20.1$ cm produces $10$ beats/s. The velocity of sound in the gas is 

speed of sound in gas speed of a travelling wave oscillation and waves waves physics

  1. $804 ms^{-1}$

  2. $402 ms^{-1}$

  3. $420 ms^{-1}$

  4. $330 ms^{-1}$

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Correct Option: B

Which relationship, out of those given below, represents the velocity of sound wave? 

$v=velocity,\ n=frequency,\ \lambda=wave\ length.$

speed of sound in gas speed of a travelling wave oscillation and waves waves physics

  1. $\displaystyle v=\frac { \lambda }{ n } $

  2. $\displaystyle v=n\lambda $

  3. $\displaystyle v=\frac { n }{ \lambda } $

  4. $\displaystyle v=n\lambda +1$

Show answer
Correct Option: B
Explanation:

Velocity of wave is equal to product of its wavelength and frequency

Newton's formula for the velocity of sound in gas is

speed of sound in gas speed of a travelling wave oscillation and waves waves physics

  1. $\displaystyle v= \sqrt {\frac {P}{\rho}}$

  2. $\displaystyle v= \frac {2}{3}\sqrt {\frac {P}{\rho}}$

  3. $\displaystyle v= \sqrt {\frac {\rho}{P}}$

  4. $\displaystyle v= \sqrt {\frac {2P}{\rho}}$

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Correct Option: A
Explanation:

Newton's formula for velocity of sound in gas is:

$\displaystyle v= \sqrt {\frac {P}{\rho}}$, where $P$ is pressure & $\rho$ is density of gas

Two monatomic ideal gases 1 and 2 of molecular masses  m$ _{1}$  and  m$ _{2}$  respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas 1 to gas 2 is given by

speed of sound in gas speed of a travelling wave oscillation and waves waves physics

  1. $\dfrac{m _{1}}{m _{2}}$

  2. $\sqrt{\dfrac{m _{1}}{m _{2}}}$

  3. $\dfrac{m _{2}}{m _{1}}$

  4. $\sqrt{\dfrac{m _{2}}{m _{1}}}$

Show answer
Correct Option: D
Explanation:
$\vartheta =\sqrt{\dfrac{\gamma RT}{M _{0}}}$

$So, \dfrac{\vartheta _{1}}{\vartheta _{2}}=\sqrt{\dfrac{\gamma RT}{M _{01}}}\times \sqrt{\dfrac{M _{02}}{\gamma RT}}$$=\sqrt{\dfrac{M _{02}}{M _{01}}}$$=\sqrt{\dfrac{m _{2}}{m _{1}}}$

Newton assumes that sound propagation in gas takes under

speed of sound in gas speed of a travelling wave oscillation and waves waves physics

  1. isothermal condition

  2. adiabatic condition

  3. isobaric condition

  4. isentropic condition

Show answer
Correct Option: A
Explanation:

Newton assumed that sound propagation in a gas takes under isothermal condition.