MHT CET · Physics · Waves and Sound
An open organ pipe and a closed organ pipe have the frequency of their first overtone identical. The ratio of length of open pipe to that of closed pipe is
- A 3:4
- B 1:2
- C 2:1
- D 4:3
Answer & Solution
Correct Answer
(D) 4:3
Step-by-step Solution
Detailed explanation
(D)
\(\frac{\ell_{0}}{\ell_{c}}=?\)
\(\begin{array}{ll}\text { For closed pipe } & n_{p}=n_{0}(2 p+1) \Rightarrow n_{l c}=n_{o}^{3} \\ \text { For open pipe } & n_{p}=n_{0}(p+1) \Rightarrow n_{l o}=n_{0}^{2}\end{array}\)
\(\begin{aligned} & n_{1 c}=3 n_{0}=\frac{3 V}{4 \ell_{c}} \\ & n_{10}=2 n_{0}=\frac{2 V}{2 \ell_{0}} \\ \therefore & \frac{3 V}{4 \ell_{c}}=\frac{2 V}{2 \ell_{0}} \\ \therefore & \frac{\ell_{0}}{\ell_{c}}=\frac{4}{3} \end{aligned}\)
\(\frac{\ell_{0}}{\ell_{c}}=?\)
\(\begin{array}{ll}\text { For closed pipe } & n_{p}=n_{0}(2 p+1) \Rightarrow n_{l c}=n_{o}^{3} \\ \text { For open pipe } & n_{p}=n_{0}(p+1) \Rightarrow n_{l o}=n_{0}^{2}\end{array}\)
\(\begin{aligned} & n_{1 c}=3 n_{0}=\frac{3 V}{4 \ell_{c}} \\ & n_{10}=2 n_{0}=\frac{2 V}{2 \ell_{0}} \\ \therefore & \frac{3 V}{4 \ell_{c}}=\frac{2 V}{2 \ell_{0}} \\ \therefore & \frac{\ell_{0}}{\ell_{c}}=\frac{4}{3} \end{aligned}\)
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