Ncert Exercises & Solutions

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,

l = \frac{λ}{2}

\Rightarrow λ = 2 l \Rightarrow \frac{c}{ν} = 2 l \Rightarrow ν = \frac{c}{2 l}

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,

ν' = \frac{c}{4 L'}

For the first harmonic,

Hence,

ν = ν'

\Rightarrow \frac{c}{2 L} = \frac{c}{4 L'}

\Rightarrow \frac{L'}{L} = \frac{2}{4} = \frac{1}{2} \Rightarrow L' = \frac{L}{2}

[ for resonance with same tuning fork]

[

∵

speed remains constant]

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