MHT CET · Physics · Waves and Sound
An open organ pipe of length ' \(l\) ' is sounded together with another open organ pipe of length \(\left(l+l_1\right)\) in their fundamental modes. Speed of sound in air is ' \(V\) '. The beat frequency heard will be ( \(l_1 \ll l\) )
- A \(\frac{\mathrm{V} l_1^2}{2 l}\)
- B \(\frac{\mathrm{V} l_1}{2 l^2}\)
- C \(\frac{\mathrm{V} l_1}{4 l^2}\)
- D \(\frac{\mathrm{V} l^2}{2 l_1}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{V} l_1}{2 l^2}\)
Step-by-step Solution
Detailed explanation
For a pipe open at both ends, \(\mathrm{f}=\frac{\mathrm{V}}{2 l}\)
\(\therefore \quad \mathrm{f}_1=\frac{\mathrm{V}}{2 l} \quad ; \quad \mathrm{f}_2=\frac{\mathrm{V}}{2\left(l+l_1\right)}\)
\(\therefore \quad\) beat frequency, \(\mathrm{f}_{\mathrm{b}}=\mathrm{f}_1-\mathrm{f}_2=\frac{\mathrm{V}}{2 l}-\frac{\mathrm{V}}{2\left(l+l_1\right)}\)
\(\therefore \quad \mathrm{f}_{\mathrm{b}}=\mathrm{V}\left[\frac{2\left(l+l_1\right)-2 l}{4 l\left(l+l_1\right)}\right]=\mathrm{V} \frac{2 l_1}{4 l\left(l+l_1\right)}\)
\(\therefore \quad \mathrm{f}_{\mathrm{b}}=\frac{\mathrm{V} l_1}{2 l\left(l+l_1\right)}\)
Since \(l_1 \ll l, l_1\) in the denominator can be taken as zero.
\(\therefore \quad \mathrm{f}_{\mathrm{b}}=\frac{\mathrm{V} l_1}{2 l^2}\)
\(\therefore \quad \mathrm{f}_1=\frac{\mathrm{V}}{2 l} \quad ; \quad \mathrm{f}_2=\frac{\mathrm{V}}{2\left(l+l_1\right)}\)
\(\therefore \quad\) beat frequency, \(\mathrm{f}_{\mathrm{b}}=\mathrm{f}_1-\mathrm{f}_2=\frac{\mathrm{V}}{2 l}-\frac{\mathrm{V}}{2\left(l+l_1\right)}\)
\(\therefore \quad \mathrm{f}_{\mathrm{b}}=\mathrm{V}\left[\frac{2\left(l+l_1\right)-2 l}{4 l\left(l+l_1\right)}\right]=\mathrm{V} \frac{2 l_1}{4 l\left(l+l_1\right)}\)
\(\therefore \quad \mathrm{f}_{\mathrm{b}}=\frac{\mathrm{V} l_1}{2 l\left(l+l_1\right)}\)
Since \(l_1 \ll l, l_1\) in the denominator can be taken as zero.
\(\therefore \quad \mathrm{f}_{\mathrm{b}}=\frac{\mathrm{V} l_1}{2 l^2}\)
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