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A closed organ pipe of length 20 cm is sounded with tuning fork in resonance. What is the frequency of tuning fork? $(v=332 \mathrm{~m} / \mathrm{s})$
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The correct answer is:
$415 Hz$
Key Idea : When length of air column is $\frac{\lambda}{4}$, then first resonance occurs.
If we adjust the length of air-column in closed organ pipe as such its any natural frequency equals to the frequency of tuning fork, then the amplitude of forced vibrations of air-column increases very much. This is the state of resonance. At first resonance
$l=\frac{\lambda}{4}$
So, frequency of tuning fork
$f=\frac{v}{\lambda}=\frac{v}{4 l}$
Given, $l=20 \mathrm{~cm}=0.2 \mathrm{~m}, v=332 \mathrm{~m} / \mathrm{s}$
Hence, $f=\frac{332}{4 \times 0.2}=415 \mathrm{~Hz}$
If we adjust the length of air-column in closed organ pipe as such its any natural frequency equals to the frequency of tuning fork, then the amplitude of forced vibrations of air-column increases very much. This is the state of resonance. At first resonance
$l=\frac{\lambda}{4}$
So, frequency of tuning fork
$f=\frac{v}{\lambda}=\frac{v}{4 l}$
Given, $l=20 \mathrm{~cm}=0.2 \mathrm{~m}, v=332 \mathrm{~m} / \mathrm{s}$
Hence, $f=\frac{332}{4 \times 0.2}=415 \mathrm{~Hz}$
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