Sounds from two identical sources S1 and S2 reach a point P. When the

1.	Sounds from two identical sources S1 and S2 reach a point P. When the sounds reach directly, and in the same phase, the intensity at P is I0. The power of S1 is now reduced by 64 %, and the phase difference between S1 and S2 is varied continuously. The maximum and minimum intensities recorded at P are now Imax and Imin.(a) Imax = 0.64 I0 (b) Imin = 0.36 I0 (c) Imax /Imin = 16 (d) Imax /Imin = 1.64/0.36
Answer» Correct Answer is: (a) I_max = 0.64 I₀ , (c) I_max /I_min = 16 Let a = initial amplitude due to S₁and S₂ each. I₀ = k(4a²), where k is a constant. After reduction of power of S₁, amplitude due to S₁ = 0.6a. Due to superposition, a_max = a + 0.6a = 1.6a, and a_min = a - 0.6a = 0.4a I_max/I_min = (a_max/a_min)² = (1.6a/0.4a) = 16.

Sounds from two identical sources S1 and S2 reach a point P. When the sounds reach directly, and in the same phase, the intensity at P is I0. The power of S1 is now reduced by 64 %, and the phase difference between S1 and S2 is varied continuously. The maximum and minimum intensities recorded at P are now Imax and Imin.(a) Imax = 0.64 I0 (b) Imin = 0.36 I0 (c) Imax /Imin = 16 (d) Imax /Imin = 1.64/0.36

Answer»

Correct Answer is: (a) I_max = 0.64 I₀ , (c) I_max /I_min = 16

Let a = initial amplitude due to S₁and S₂ each.

I₀ = k(4a²), where k is a constant.

After reduction of power of S₁, amplitude due to S₁ = 0.6a.

Due to superposition, a_max = a + 0.6a = 1.6a, and

a_min = a - 0.6a = 0.4a

I_max/I_min = (a_max/a_min)² = (1.6a/0.4a) = 16.

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