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 $y_1$ = $Asin\left(\frac{\pi x}{L}\right)sin\omega t$ and energy is $E_1$ and in another experiment its displacement is $y_2$ = $Asin\left(\frac{2\pi x}{L}\right)sin2\omega t$ and energy is $E_2$. Then

• A. E₂ = E₁
• B. E₂ = 2E₁
• C. E₂ = 4E₁
• D. E₂ = 16E₁

Answer 1

The correct option is C

E₂ = 4E₁

Energy $E$ $\propto$ ${\left(amplitude\right)}^2$${\left(frequency\right)}^2$

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

Therefore,$E_2$ = $4E_1$