Two sound waves move in the same diretion 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 saection 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 for a given medium, velocity of sound v and the bulk modulus B are constant, I depends only on the pressure amplitude. Since in the given problem pressure amplitude of both waves are same,

hence $I_{1} = I_{2}$
an $P_{1} = P_{2}$

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Q. 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

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Two sound waves move in the same diretion 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 saection by the first wave be P1 and that by the second wave be P2 . Then

The correct option is A P1=P2Since for a given medium, velocity of sound v and the bulk modulus B are constant, I depends only on the pressure amplitude. Since in the given problem pressure amplitude of both waves are same, hence I1=I2 an P1=P2

The correct option is A $P_{1} = P_{2}$
Since for a given medium, velocity of sound v and the bulk modulus B are constant, I depends only on the pressure amplitude. Since in the given problem pressure amplitude of both waves are same,

hence $I_{1} = I_{2}$
an $P_{1} = P_{2}$