Question

The ends of a stretched wire of length L are fixed at $x = 0$ and $x = L$ . In one experiment, the displacement of the wire is ${y_1} = A\sin (2\pi x/L)\sin \omega t$ and energy is ${E_1}$ and in another experiment its displacement is ${y_2} = A\sin (2\pi x/L)\sin 2\omega t$ and energy is ${E_2}$ . Then.
(A) ${E_1} = {E_2}$
(B) ${E_2} = 2{E_1}$
(C) ${E_2} = 4{E_1}$
(D) ${E_2} = 16{E_1}$

Answer 1

Hint : To solve this question, we have to know about kinetic energy. A force must be applied on a body to accelerate an object. Work must be done in order to apply a force. The body will move with an unvarying speed after the work has been done due to the energy provided by it. The speed and the mass of the body are factors on which the energy transfer that makes up the kinetic energy depends.

Complete Step By Step Answer:
We know that,
$\int d {E_k} = \int\limits_0^1 {\left( {\dfrac{1}{2}\mu {A^2}{\omega ^2}{{\sin }^2}\dfrac{{2\pi x}}{L}} \right)} dx$
$= \dfrac{{\mu {A^2}{\omega ^2}L}}{4}$
Total energy is equal to total kinetic energy $+$ total potential energy
Let us consider, the total energy is ${E_1}$
${E_1} = \dfrac{{\mu {A^2}{\omega ^2}L}}{4} + \dfrac{{\mu {A^2}{\omega ^2}L}}{4}$
$= \dfrac{{\mu {A^2}{\omega ^2}L}}{2}$
Similarly, by the second wave, let us consider the total energy as ${E_2}$
Which is $= \dfrac{{\mu {A^2}{{(2\omega )}^2}L}}{2}$
As the frequency is double, that is $4{E_1}$
$\therefore {E_2} = 4{E_1}$
So, the correct option is option number C.

Note :
We also have to know that the formula for potential energy depends on the force acting on the two objects. For the gravitational force the formula is: $W = mgh$ . Where, m is the mass in kilograms. g is the acceleration due to gravity. h is the height in meters. We know that, Potential energy is the energy that is stored in an object due to its position relative to some zero position. We have to keep that in our mind.