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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 y1 = A sin(πx/L) sin ωt and the energy is E1, and in another experiment the displacement is y2 = A sin(2πx/L) sin 2ωt and the energy is E2. Then(a) E2 = E1 (b) E2 = 2E1 (c) E2 = 4E1 (d) E2 = 16E1 |
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Answer» Correct Answer is: (c) E2 = 4E1 A stationary wave has the equation of the form y = A sin kx sin ωt. Here, for y1, we have k1 = π/L, ω1 = ω. ∴ v1 = ω1/k1 = ωL/π. For y2, we have k2 = 2π/L, ω2 = 2ω. ∴ v2 = ω2/k2 = ωL/π = v1. Thus, the wave velocities are the same in both cases. Also, they have the same amplitude. The frequency for y2 is twice the frequency for y1. Now, B ∵ energy ∝ (frequency)2 , ∴ E2 = 4E1. |
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