A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP in 300 m/s, the frequency difference between the fundamental and second harmonic of this pipe is ________ Hz.Correct answer is '106'. Can you explain this answer?

Question

Answer

Frequency of the fundamental mode of vibration in an open organ pipe can be given by the formula:

f1 = (v / 2L)

where f1 is the frequency of the fundamental mode, v is the speed of sound, and L is the length of the pipe.

Similarly, the frequency of the second harmonic can be given by:

f2 = (2v / 2L) = (v / L)

Given that the length of the pipe is 1 meter and the speed of sound in air at STP is 300 m/s, we can calculate the frequencies as follows:

f1 = (300 / 2*1) = 150 Hz
f2 = (300 / 1) = 300 Hz

The frequency difference between the fundamental and second harmonic is given by:

Δf = f2 - f1

Substituting the values, we get:

Δf = 300 - 150 = 150 Hz

However, in this case, the organ pipe is not in air, but in a gas that has double the density of air at STP. Since the speed of sound is inversely proportional to the square root of the density of the medium, the speed of sound in this gas would be:

v' = v / sqrt(2)

Using this new speed of sound, we can calculate the frequencies as follows:

f1' = (v' / 2L) = (v / sqrt(2)) / 2L = (300 / sqrt(2)) / 2*1 = 106.07 Hz
f2' = (v' / L) = (v / sqrt(2)) / L = (300 / sqrt(2)) / 1 = 212.14 Hz

The frequency difference between the fundamental and second harmonic in this case is:

Δf' = f2' - f1' = 212.14 - 106.07 = 106.07 Hz

Therefore, the correct answer is 106 Hz.

Answer

v_air = 300 m/s

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