Question

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=4P.

Answer 1

Option (a) is Correct

Explanation:

a) $P_1 = P_2$

As  

Since the average power transmitted by a wave is independent of the wavelength,  

we have $P_1 = P_2.$

Answer 2

Given:

Sound waves = 2

First wave = P1

Second Wave = P2

To Find:

Relationship between the waves

Solution:

Let the cross-sectional area be = x

The intensity of first wave = I1

The intensity of second wave = I2

Therefore,

P₁ = xI1  and P2 = xi2.

But intensity I =P0²V/2B

Since, for the given medium of sound velocity (V) and bulk modulus (B) are constant, thus I will be dependent only on the pressure amplitude P0.

Now, as the pressure amplitude of both waves are the same, thus I1 - I2.

Answer: When two sound waves move in the same direction and in the same medium where the pressure amplitudes are also equal then P1 = P2.