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 and energy is E1, and in another experiment its displacement is and energy is E2. Then
(1) E2 = E1
(2) E2 = 2E1
(3) E2 = 4E1
(4) E2 = 16E1
(3) Energy (E) ∝ (Amplitude)2 (Frequency)2
Amplitude is same in both the cases, but frequency 2ω in the second case is two times the frequency (ω) in the first case. Hence E2 = 4E1.