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Question: The ends of a stretched wire of length \(L\) are fixed at \(x = 0\) and \(x = L\). In one experiment...

The ends of a stretched wire of length LL are fixed at x=0x = 0 and x=Lx = L. In one experiment, the displacement of the wire is y1=Asin⁡(πxL)sin⁡ωt{y_1} = A\sin (\dfrac{{\pi x}} {L})\sin \omega t and energy is E1{E_1} and in another experiment its displacement is y2=Asin⁡(2πxL)sin⁡2ωt{y_2} = A\sin (\dfrac{{2\pi x}} {L})\sin 2\omega t and energy is E2{E_2} . Then:
A) E2=E1{E_2} = {E_1}
B) E2=2E1{E_2} = 2{E_1}
C) E2=4E1{E_2} = 4{E_1}
D) E2=16E1{E_2} = 16{E_1}

Explanation

Solution

Total energy for any stretched wire can be defined as sum of kinetic energy and potential energy. Think of the formula you can use to calculate kinetic energy involving the displacement of the wire.

Complete step by step solution:
Elastic potential energy is stored by application of an elastic object deformation force. The energy is stored until the force is removed and the object returns to its original shape. Deformation might involve the object being compressed, stretched or twisted. An object designed to store elastic energy potentially has a high elastic limit, but any elastic object has a load limit. The object will no longer return to its original form when it is deformed beyond its elastic limit. Wind-up mechanical clocks powered by spindles have been popular supplies in earlier generations. Elastic materials, such as rubber strips and plastics, can function as springs but are frequently hysteresis. That means that when the material is deformed, the force-extending curve follows a different path compared to when relaxed back into its balance position. Now, in order to calculate the total energy of the first, we need to calculate the kinetic energy first, which can be calculated by,
∫dEk=∫0112μA2ω2sin22πxLdx\smallint d {E_k} = \int\limits_0^1 {\dfrac{1}{2}} \mu {A^2} {\omega ^2} si {n^2}\dfrac {{2\pi x}} {L} dx
Where,
Ek{E_k} is the kinetic energy of the first wave.
=μA2ω2L4= \dfrac{{\mu {A^2} {\omega ^2} L}} {4}
Now,
The total energy for the first wave would be equal to sum of kinetic energy and potential energy,
E1=μA2ω2L4+μA2ω2L4{E_1} = \dfrac{{\mu {A^2} {\omega ^2} L}} {4} + \dfrac{{\mu {A^2} {\omega ^2} L}} {4}
E1=μA2ω2L2{E_1} = \dfrac{{\mu {A^2} {\omega ^2} L}} {2}
Now,
The total energy for the second wave would be,
E2=μA2(2ω)2L2{E_2} = \dfrac{{\mu {A^2} {{(2\omega)} ^2} L}} {2}
Hence, from the above equation,
E2=4E1{E_2} = 4{E_1}
Therefore, the correct option is C.

Note: Fortunately, also for elastic substances in general, the basic technique of the definition of work which we employed in an optimal spring. In the region under the force vs. the extensions curve, the elastic potential energy is always found regardless of the curve form.