A tuning fork is vibrating at 250 Hz. The length of the shortest closest organ pipe that will resonate with the tuning fork will be_____cm.

(Take speed of sound in air as 340 ms^-1)

$\frac{\lambda }{4}=\mathcal{l}\Rightarrow \lambda =4\mathcal{l}$

$f=\frac{v}{\lambda }=\frac{v}{4\mathcal{l}}\Rightarrow \mathcal{l}=\frac{v}{4f}$

$\mathcal{l}=\frac{34}{4\times 25}=0.34m=34cm$

 

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