Two sound waves move in the same direction in the same medium. The pressure amplitudes of the waves are equal but the wavelength,

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Two sound waves move in the same direction in the same medium. The pressure amplitudes of the waves are equal but the wavelength,

Two sound waves move in the same direction in the same medium. The pressure amplitudes of the waves are equal but the wavelength of the first wave is double the second. Let the average power transmitted across a cross-section by the first wave be P₁ and that by the second wave be P₂ . Then

(a) P₁= P₂

(b) P₁= 4P₂

(c) P₂= 2P₁

(d) P₂ = 4P₁.

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Salim Ahmed Kawsar · Answer 1 · Feb 05, 2023 15:29:48

The correct answer is (a) P₁= P₂

EXPLANATION:

Let the cross-section area = A, if the intensity of the first wave at this cross-section = I₁ and of the second wave = I₂, then P₁ =AI₁ and P₂ = AI₂. But the intensity I =p₀²V/2B

Since for a given medium velocity of sound V and the bulk modulus B are constant, I depends only on the pressure amplitude p₀. Since in the given problem pressure amplitude of both waves are same hence I₁ = I₂

→P₁ = P₂

So the option (a).

Two sound waves move in the same direction in the same medium. The pressure amplitudes of the waves are equal but the wavelength,

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