A) 1250 Hz
B) 625 Hz
C) 417 Hz
D) 715 Hz
Correct Answer: B
Solution :
In a closed organ pipe in which length of air-column can be increased or decreased. The first resonance occrus at \[\lambda /4\] and second resonance occurs at \[3\lambda /4\]. Thus, at first resonance \[\frac{\lambda }{4}=13\] ... (i) and at second resonance \[\frac{3\lambda }{4}=41\] ... (ii) Subtracting Eq. (i) from Eq. (ii), we have \[\frac{3\lambda }{4}-\frac{\lambda }{4}=41-13\] \[\Rightarrow \] \[\frac{\lambda }{2}=28\] \[\therefore \] \[\lambda =56\,\,cm\] Hence, frequency of tuning fork \[n=\frac{v}{\lambda }=\frac{350}{56\times {{10}^{-2}}}=625\,Hz\] Note: In resonance tube antinode is not formed exactly at open end but it is formed a little above the open end known as end correction [e]. So. in first and second state of air-column, the lengths are \[{{l}_{1}}+e\] and \[{{l}_{2}}+e\].You need to login to perform this action.
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