A closed pipe is suddenly opened and changed to an open pipe of same length. The fundamental frequency of the resulting open pipe is less than that of 3rd harmonic of the earlier closed pipe by 55 Hz. Then the value of fundamental frequency of the closed pipe is

165 Hz

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100 Hz

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55 Hz

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220 Hz

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Solution

The correct option is C 55 Hz
Frequency of first mode of vibration

$n_{1} = \frac{3 v}{4 l} ⟶ (1)$ (closed pipe)

and frequency of first mode of vibration

$n_{2} = \frac{v}{2 l} ⟶ (2)$ (open pipe)

According to question,

$n_{1} - n_{2} = \frac{3 v}{4 l} - \frac{v}{2 l} = 55 o r, \frac{3 v - 2 v}{4 l} = 55 ∴ \frac{v}{4 l} = 55 H z$

Which is fundamental frequency of the closed pipe

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A closed pipe is suddenly opened and changed to an open pipe of same length. The fundamental frequency of the resulting open pipe is less than that of 3rd harmonic of the earlier closed pipe by 55 Hz. Then the value of fundamental frequency of the closed pipe is

The correct option is C 55 HzFrequency of first mode of vibration n1=3v4l⟶(1) (closed pipe) and frequency of first mode of vibration n2=v2l⟶(2)(open pipe) According to question, n1−n2=3v4l−v2l=55or,3v−2v4l=55∴v4l=55Hz Which is fundamental frequency of the closed pipe