Question:
A tuning fork is vibrating at $250 \mathrm{~Hz}$. The length of the shortest closed organ pipe that will resonate with the tuning fork will be $\mathrm{cm}$.
(Take speed of sound in air as $340 \mathrm{~ms}^{-1}$ )
Solution:
$\frac{\lambda}{4}=\ell \Rightarrow \lambda=4 \ell$
$\mathrm{f}=\frac{\mathrm{V}}{\lambda}=\frac{\mathrm{V}}{4 \ell}$
$\Rightarrow 250=\frac{340}{4 \ell}$
$\Rightarrow \ell=\frac{34}{4 \times 25}=0.34 \mathrm{~m}$
$\ell=34 \mathrm{~cm}$
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