An organ pipe of length L open at both ends is found to vibrate in its first harmonic when sounded with a tuning fork of 480 Hz. What should be the length of a pipe closed at one end, so that it also vibrates in its first harmonic with the same tuning fork?

Hint: Use the concept of standing waves.
Step 1: Find the frequency of the open pipe.
Consider the situation in the diagram. As the organ pipe is open at both ends. Hence, for the first harmonic,
$\mathrm{l}=\frac{\mathrm{\lambda }}{2}$

where c is the speed of the sound wave in the air.
Step 2: Find the frequency of the closed pipe.
For pipe closed at one end, $\mathrm{\nu }\text{'}=\frac{\mathrm{c}}{4\mathrm{L}\text{'}}$
For the first harmonic, Hence,
$\mathrm{\nu }=\mathrm{\nu }\text{'}$
$⇒\frac{\mathrm{c}}{2\mathrm{L}}=\frac{\mathrm{c}}{4\mathrm{L}\text{'}}$
$⇒\frac{\mathrm{L}\text{'}}{\mathrm{L}}=\frac{2}{4}=\frac{1}{2}⇒\mathrm{L}\text{'}=\frac{\mathrm{L}}{2}$                                                      [ for resonance with same tuning fork]
[$\because$ speed remains constant]