The fundamental frequency of an open pipe is | vedclass

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Question

(D) The fundamental frequency of an open pipe is given by $n = \frac{v}{2 \ell}$.For a closed pipe, the harmonics are odd multiples of the fundamental

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(D) The fundamental frequency of an open pipe is given by $n = \frac{v}{2 \ell}$.For a closed pipe, the harmonics are odd multiples of the fundamental frequency $n_0 = \frac{v}{4 \ell}$.The $2^{nd}$ harmonic of a closed pipe is actually the $1^{st}$ overtone, which is the $3^{rd}$ harmonic, given by $3 n_0 = \frac{3v}{4 \ell}$.According to the problem, the frequency of the $2^{nd}$ harmonic of the closed pipe is $200 \,Hz$ higher than $n$:$\frac{3v}{4 \ell} - n = 200$Since $n = \frac{v}{2 \ell}$, we have $\frac{v}{4 \ell} = \frac{n}{2}$.Substituting this into the equation: $3(\frac{n}{2}) - n = 200$.$1.5n - n = 200 \implies 0.5n = 200$.$n = 400 \,Hz$.

The fundamental frequency of an open pipe is $n$. If it is closed from one end, then the frequency of the $2^{nd}$ harmonic of the closed pipe is higher by $200 \,Hz$ than $n$. The value of $n$ is: (in $\,Hz$)

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