Question

For fundamental frequency $f$ of a closed pipe, choose the correct options.

• A. If radius of pipe is increased $f$ will decrease
• B. If temperature is increased $f$ will increase
• C. If molecular mass of the gas filled in the pipe is increased $f$ will decrease.
• D. If pressure of gas (filled in the pipe) is increased without change in temperature, $f$ will remain unchanged

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A If radius of pipe is increased $f$ will decrease
B If temperature is increased $f$ will increase
C If molecular mass of the gas filled in the pipe is increased $f$ will decrease.
D If pressure of gas (filled in the pipe) is increased without change in temperature, $f$ will remain unchanged

Fundamental frequency of a closed pipe with end correction: ${\nu }_1=\frac{v}{4(L+0.6r)}$

where $r$ and $L$ be the radius and length of the pipe respectively.

Thus when radius is decreased then frequency will increase.

Velocity of the waves $v=\sqrt{\frac{\gamma P}{\rho }}ORv=\sqrt{\frac{\gamma RT}{M}}$ ...............(1)

This stats that if temperature is increased, then velocity will increase and hence frequency will also increase.

Similarly $f\propto \sqrt{\frac{1}{M}}$, thus frequency will decrease when molecular mass is increased.

Also when pressure is increased keeping the temperature constant, then frequency will remain the same. (At constant temperature, changing pressure has no effect on frequency because density will also change accordingly.)