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
$P_{1} = P_{2}$

Power = $\frac{P_{0}^{2} V (a r e a)}{2 B}$

Velocity is same as mediums are same

So B also same

Power $α P_{0}^{2}$

Since pressure amplitude is also same so

$P_{1} = P_{2}$

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Similar questions

Two sound waves move in the same direction. If the average power transmitted across a cross-section by them are equal while their wavelengths are in the ratio of 1:2. Their pressure amplitudes would be in the ratio of ___.

Q. The displacement function

S (x, t)

of three sound waves

1, 2

and

3

at a certain time

t

as marked on the curves that travel along

x - axis

through air. If

P_{1}, P_{2} a n d P_{3}

represent their pressure amplitudes respectively, then the correct relation between them is:

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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 = P2 Power =P20V(area)2B Velocity is same as mediums are same So B also same Power α P20 Since pressure amplitude is also same so P1 = P2

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

Power = $\frac{P_{0}^{2} V (a r e a)}{2 B}$

Velocity is same as mediums are same

So B also same

Power $α P_{0}^{2}$

Since pressure amplitude is also same so

$P_{1} = P_{2}$