Energy in wave motion - class-XI

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 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

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

The correct option is (b)

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.

For a travelling wave y = f(x - vt), the energy density is related to the square of the partial derivative of the wave function with respect to time, specifically (mu/2) * (dy/dt)^2.

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)

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 the extent of stretching.

The correct option is (c)

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 $

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

 Here, we will take ,

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. 

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

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, $u _1:u _2=25:4$

The kinetic energy passing through a point is calculated by integrating the power or using the energy density formula. Given the parameters, the calculation leads to 9.

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.