Two sound waves move in the same direction in the same medium. The pressure amplitude of the waves are equal but the wavelength of the first wave is double that of the second. Let the average power transmitted across a cross section by the two waves be $P_{1}$ and $P_{2}$

P_{1} / P_{2}

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P_{1} / P_{2} = 2

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s_{1} / s_{2} = 1 / 2

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s_{1} / s_{2} = 2 / 1

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Solution

The correct option is
A
$P_{1} / P_{2} = 1$

Given, $A_{1} = A_{2}$
$λ_{1} = 2 λ_{2}$
$P = \frac{1}{2} μ W^{2} A^{2} V$
$∴ \frac{P_{1}}{P_{2}} = \frac{1 / 2 \times μ W^{2} A_{1}^{2} V}{1 / 2 \times μ W^{2} A_{2}^{2} V} = 1$
$\Rightarrow P_{1} = P_{2}$

Hence Option (A) is correct.

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

The correct option is A P1/P2=1Given, A1=A2 λ1=2λ2P=12μW2A2V∴P1P2=1/2×μW2A21V1/2×μW2A22V=1⇒P1=P2Hence Option (A) is correct.

Two sound waves move in the same direction in the same medium. The pressure amplitude of the waves are equal but the wavelength of the first wave is double that of the second. Let the average power transmitted across a cross section by the two waves be $P_{1}$ and $P_{2}$

The correct option is
A
$P_{1} / P_{2} = 1$

Given, $A_{1} = A_{2}$
$λ_{1} = 2 λ_{2}$
$P = \frac{1}{2} μ W^{2} A^{2} V$
$∴ \frac{P_{1}}{P_{2}} = \frac{1 / 2 \times μ W^{2} A_{1}^{2} V}{1 / 2 \times μ W^{2} A_{2}^{2} V} = 1$
$\Rightarrow P_{1} = P_{2}$

Hence Option (A) is correct.