Tag: waves

Questions Related to waves

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 $

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.