Tag: energy in wave motion

Questions Related to energy in wave motion

If the energy density and velocity of a wave are $u$ and $c$ respectively then the energy propagating per second per unit area will be

energy in wave motion oscillation and waves waves physics

  1. $u/c$

  2. $c^2u$

  3. $uc$

  4. $c/u$

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Correct Option: C
Explanation:

If the energy density and velocity of a wave are $u$ and $c$ respectively then the energy propagating per second per unit area will be $uc.$

The kinetic energy per unit length for a wave on a string is the positional coordinate

energy in wave motion oscillation and waves waves physics

  1. True

  2. False

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Correct Option: B
Explanation:

The kinetic energy for a particle is given by $(\mu \Delta x/ 2) (\dfrac{dy}{dt})^2$

Thus, it depends only on the time variable and not on the position variable

A travelling wave has an equation of the form $A(x,t)=f(x+vt)$. The relation connecting positional derivative with time derivative of the function is:

energy in wave motion oscillation and waves waves physics

  1. $\dfrac{dA}{dt}=\pm v^2 \dfrac {dA}{dx}$

  2. $\dfrac{dA}{dt}=\pm v \dfrac {dA}{dx}$

  3. $\dfrac{dA}{dt}=\pm \sqrt(v) \dfrac {dA}{dx}$

  4. $\dfrac{dA}{dt}=(2 \pi v/\lambda) \dfrac {dA}{dx}$

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Correct Option: B
Explanation:

Positional derivative and time derivative of a function f is $\dfrac{dA}{dt}=\pm v \dfrac {dA}{dx}$

The correct option is (b)

Kinetic energy per unit length for a particle in a standing wave is zero at:

energy in wave motion oscillation and waves waves physics

  1. nodes

  2. antinodes

  3. mid-way between a node and an antinode

  4. None of the above

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Correct Option: B
Explanation:

Particle at antinodes is momentarily at rest and hence has zero kinetic energy. Its speed comes down to zero at this point and all energy is stored in the form of potential energy.

The total energy per unit length for a travelling wave in a string of mass density $\mu$ , whose wave function is $A(x,t) = f(x \pm vt)$ is given by: 

energy in wave motion oscillation and waves waves physics

  1. $E _tot = \sqrt(\mu/2) (\dfrac{dA}{dt})^2$

  2. $E _tot = (\mu/2) (\dfrac{dA}{dt})^2$

  3. $E _tot = (\mu/2)^2 (\dfrac{dA}{dt})^2$

  4. $E _tot = (2\mu) (\dfrac{dA}{dt})^2$

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Correct Option: B

The maximum potential energy / length increases with:

energy in wave motion oscillation and waves waves physics

  1. Amplitude

  2. Wavelength

  3. Frequency

  4. Velocity

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Correct Option: C
Explanation:

If $y= A \sin (\omega t- kx)$ is the equation of a wave through a string, then the slope of the wave is $\dfrac{dy}{dx}=- Ak \cos (\omega t- kx)$. 

The maximum potential energy will be $T \times \Delta x \times (\dfrac{dy}{dx})^2A^2k^2=A^2k^2 \cos ^2(\omega t -  kx)T \Delta x$

The maximum potential energy will be obtained if cos (\omega t - kx)=1. Thus, maximum potential energy = $4 \pi^2A^2T f \times T \times \Delta x $; T is the tension in the string

We also know that $T=\mu v^2$. Substituting, we get, 

Maximum potential energy = $4 \pi^2 f^2A^2 \mu$
Thus maximum potential energy depends on frequency and as frequency increases, potential energy also increases

The correct option is (c)

In the absence of a wave travelling on a tight rope fixed at both the ends, the potential energy per unit length on the rope is zero.

energy in wave motion oscillation and waves waves physics

  1. True

  2. False

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Correct Option: B
Explanation:

Potential energy of a string depends only on how much the string is stretchable and not on the wave passing through it. A wave passing through it can provide some energy, thereby stretching it

In the problem given, the string is already stretched by keeping it between two supports. Thus, potential energy is not zero

Potential energy of a string depends on 

energy in wave motion oscillation and waves waves physics

  1. Wave velocity

  2. Amplitude of the wave

  3. Extent of stretching of the string

  4. None of the above

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Correct Option: C
Explanation:

Potential energy of a string depends on the extent of stretching.

The correct option is (c)

The wave number of energy emitted when electron jumps from fourth orbit to seconds orbit in hydrohen in $20,497\ cm^{-1}$. The wave number of energy for the same transition in $He^{+}$ is

energy in wave motion oscillation and waves waves physics

  1. $5,099\ cm^{-1}$

  2. $20,497\ cm^{-1}$

  3. $40,994\ cm^{-1}$

  4. $81,988\ cm^{-1}$

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Correct Option: D
Explanation:

Given that,

Wave number of energy for hydrogen $=20497\,c{{m}^{-1}}$

Now, for hydrogen

$\dfrac{1}{{{\lambda } _{H}}}=R\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{4}^{2}}} \right)=20497$

Now, for helium

  $ \dfrac{1}{\lambda }=R\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{4}^{2}}} \right)\times {{2}^{2}} $

 $ \dfrac{1}{{{\lambda } _{He}}}=20497\times 4 $

 $ \dfrac{1}{{{\lambda } _{He}}}=81988\,c{{m}^{-1}} $

Hence, the wave number of energy for helium is $81988\ cm^{-1}$

The ends of a stretched string of length $L$ are fixed at $x=0$ and $x=L$. In one experiment, the displacement of the wire is $y _{1}=2A\sin\left(\dfrac{\pi x}{L}\right)\sin\omega t$ and energy $E _1$ and in another experiment, its displacement is $y _2 = A\sin\left({\displaystyle\frac{2\pi x}{L}}\right)\sin{2\omega t}$ and energy $E _2$ then

energy in wave motion oscillation and waves waves physics

  1. $E _2 = E _1$

  2. $E _2 = 2E _1$

  3. $E _2 = 4E _1$

  4. $ 16E _1$

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Correct Option: C
Explanation:

As $K _1 = \dfrac{\pi}{L}$ and $K _2=\dfrac{2\pi}{L}$ and $\omega _2 = 2\omega _1 $ , 
Wave velocity $v=\dfrac{\omega}{K}$ is same for both the cases as well as amplitudes in both the cases . 
Hence , Energy $\propto f^2 \rightarrow $ Energy $\propto \omega ^2  $
$E _2 = 4 E _1 $