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- Asin - x - L sin - t and energy is E 1- and in another experiment its displacement is y 2- A sin2 - x - L sin 2 - t and energy is E 2- ThenA- E 2- E 1B- E 2-2 E 1C- E 2-4 E 1D- E 2-16 E 1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Transmission of Wave

The ends of a...

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 $y_{1} = A s i n (\frac{π x}{L}) s i n ω t$ and energy is $E_{1}$ , and in another experiment its displacement is $y_{2} = A s i n (\frac{2 π x}{L}) s i n 2 ω t$ and energy is $E_{2}$ . Then

E₂=E₁

No worries! We‘ve got your back. Try BYJU‘S free classes today!

E₂=2E₁

No worries! We‘ve got your back. Try BYJU‘S free classes today!

E₂=4E₁

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

E₂=16E₁

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
E₂=4E₁

$E n e r g y (E) \propto (A m p l i t u d e)^{2} (F r e q u e n c y)^{2}$

Amplitude is same in both the cases, but frquency $2 ω$ in the second case is two times the frequency $(ω)$ in the first case.

Hence $E_{2} = 4 E_{1}$

Suggest Corrections

Similar questions

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 $y_{1}$ = $A s i n (\frac{π x}{L}) s i n ω t$ and energy is $E_{1}$ and in another experiment its displacement is $y_{2}$ = $A s i n (\frac{2 π x}{L}) s i n 2 ω t$ and energy is $E_{2}$ . Then