An air column in a pipe which is closed at on | vedclass

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(B) For a pipe closed at one end, the resonance occurs when the length of the air column $L$ is an odd multiple of a quarter of the wavelength $\lambd

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) For a pipe closed at one end, the resonance occurs when the length of the air column $L$ is an odd multiple of a quarter of the wavelength $\lambda$.The condition for resonance is $L = (2n - 1) \frac{\lambda}{4}$, where $n = 1, 2, 3, ...$First, calculate the wavelength $\lambda$ using the formula $v = f \lambda$, where $v = 332$ m/s and $f = 415$ Hz.$\lambda = \frac{v}{f} = \frac{332}{415} = 0.8$ m = $80$ cm.Now, substitute $\lambda = 80$ cm into the resonance condition:$L = (2n - 1) \frac{80}{4} = (2n - 1) 	imes 20$ cm.For $n = 1, L = 20$ cm.For $n = 2, L = 60$ cm.For $n = 3, L = 100$ cm.Comparing these values with the given options, $40$ cm is not an odd multiple of $20$ cm, therefore it is not in resonance.

An air column in a pipe which is closed at one end will be in resonance with a vibrating tuning fork of frequency $415$ Hz for various vibrating air columns. Which one of the following lengths is not in resonance (in $cm$)? (Velocity of sound in air = $332$ m/s) (Neglect end correction)

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