A stationary tuning fork is in resonance with an air column in a pipe. If the tuning fork is moved with a speed of 2 in front of the open end of the pipe and parallel to it, the length of the pipe should be changed for the resonance to occur with the mov

Question

A stationary tuning fork is in resonance with an air column in a pipe. If the tuning fork is moved with a speed of 2 ${ms}^{- 1}$ in front of the open end of the pipe and parallel to it, the length of the pipe should be changed for the resonance to occur with the moving tuning fork. If the speed of sound in air is 320 ${ms}^{- 1}$ , the smallest value of the percentage change required in the length of the pipe is [2020]

Smart Achievers · Accepted Answer

(0.62)

Let $ℓ_{1} = initial length of pipe$

$ℓ_{2} = new length of pipe$

$V_{T} = Speed of tuning fork$

In closed organ pipe, $f = \frac{V}{4 ℓ_{1}}$

When tuning fork is moved, $f' = f (\frac{V}{V - V_{T}}) = \frac{V}{4 ℓ_{2}}$

$\Rightarrow \frac{V}{4 ℓ_{1}} (\frac{V}{V - V_{T}}) = \frac{V}{4 ℓ_{2}}$ $\Rightarrow \frac{V - V_{T}}{V} = \frac{ℓ_{2}}{ℓ_{1}}$

$\Rightarrow \frac{ℓ_{2}}{ℓ_{1}} - 1 = \frac{V - V_{T}}{V} - 1$ $\Rightarrow \frac{ℓ_{2} - ℓ_{1}}{ℓ_{1}} = - \frac{V_{T}}{V}$

Percentage change required in the length of the pipe,

$\frac{ℓ_{2} - ℓ_{1}}{ℓ_{1}} \times 100 = \frac{- 2}{320} \times 100 = - 0.625 %$

Hence, smallest value of percentage change required in the length of pipe is 0.625%

Solution

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