MHT CET · Physics · Waves and Sound
If ' \(l\) ' is the length of the open pipe, ' \(r\) ' is the internal radius of the pipe and ' \(V\) ' is the velocity of sound in air then fundamental frequency of open pipe is
- A \(\frac{\mathrm{V}}{(l+0.3 \mathrm{r})}\)
- B \(\frac{\mathrm{V}}{(l+1.2 \mathrm{r})}\)
- C \(\frac{\mathrm{V}}{(l+0.6 \mathrm{r})}\)
- D \(\frac{\mathrm{V}}{2(l+1.2 \mathrm{r})}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{V}}{2(l+1.2 \mathrm{r})}\)
Step-by-step Solution
Detailed explanation
For an open organ pipe, the length of the pipe with end correction is given as:
\(
\begin{aligned}
& \mathrm{L}=l+2 \mathrm{e}=l+2 \times 0.6 \mathrm{r} \\
& \mathrm{L}=l+1.2 \mathrm{r}
\end{aligned}
\)
\(\therefore \quad\) The fundamental frequency of open pipe is:
\(
\begin{aligned}
& \mathrm{f}=\frac{\mathrm{v}}{2 \mathrm{~L}} \\
& \mathrm{f}=\frac{\mathrm{v}}{2(l+1.2 \mathrm{r})}
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{L}=l+2 \mathrm{e}=l+2 \times 0.6 \mathrm{r} \\
& \mathrm{L}=l+1.2 \mathrm{r}
\end{aligned}
\)
\(\therefore \quad\) The fundamental frequency of open pipe is:
\(
\begin{aligned}
& \mathrm{f}=\frac{\mathrm{v}}{2 \mathrm{~L}} \\
& \mathrm{f}=\frac{\mathrm{v}}{2(l+1.2 \mathrm{r})}
\end{aligned}
\)
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