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 $P_{1}$ and that by the second wave be $P_{2} .$ Then

P_{1} = P_{2}

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P_{1} = {4 P}_{2}

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P_{2} = {2 P}_{1}

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P_{2} = {4 P}_{1}

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Solution

The correct option is A $P_{1} = P_{2}$
Since the sound waves are travelling in the same direction and in the same medium with equal amplitude, then the power will also be equal, i.e., $P_{1} = P_{2}$ .

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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 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

The correct option is A P1=P2Since the sound waves are travelling in the same direction and in the same medium with equal amplitude, then the power will also be equal, i.e., P1=P2.

The correct option is A $P_{1} = P_{2}$
Since the sound waves are travelling in the same direction and in the same medium with equal amplitude, then the power will also be equal, i.e., $P_{1} = P_{2}$ .