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An open pipe is in resonance in its 2nd harmonic with tuning fork of frequency f_1. Now, it's closed at one end. If the frequency of the tuning fork is increased slowly from f_1 then again a resonance is obtained with a frequency f_2. If in this case the pipe vibrates with nth harmonic, then-

Option: 1

n=3, f_2=\frac{7}{4} f_1


Option: 2

n=3, f_2=\frac{5}{4} f_1


Option: 3

f2>f1\hspace{2mm}or\hspace{2mm} f2=\frac{h}{4}f1


Option: 4

n=5, f_2=\frac{3}{4} f_1


Answers (1)

best_answer

\begin{aligned} & f_1=2\left(\frac{v}{2 l}\right)=\frac{v}{l} \\ & f_2=n\left(\frac{v}{2 l}\right) \\ & f_2>f_1 \text { or } f_2=\frac{h}{4} \cdot f_1 \end{aligned}

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Rakesh

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