Question:
Two open organ pipes of lengths \(50\,\text{cm}\) and \(51\,\text{cm}\) are totally immersed in a medium. They are found to give \(40\) beats in \(10\,\text{s}\) when each is sounding at its fundamental note. The speed of sound in this medium is
Two open organ pipes of lengths \(50\,\text{cm}\) and \(51\,\text{cm}\) are totally immersed in a medium. They are found to give \(40\) beats in \(10\,\text{s}\) when each is sounding at its fundamental note. The speed of sound in this medium is
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For an open organ pipe, the fundamental frequency is \(f=\frac{v}{2L}\). Beat frequency is the difference between the two frequencies.
Updated On: Jun 26, 2026
- \(275\,\text{ms}^{-1}\)
- \(310\,\text{ms}^{-1}\)
- \(258\,\text{ms}^{-1}\)
- \(204\,\text{ms}^{-1}\)
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The Correct Option is D
Solution and Explanation
Step 1: Write the fundamental frequency of an open organ pipe.
For an open organ pipe, the fundamental frequency is \[ f=\frac{v}{2L} \] where \(v\) is the speed of sound and \(L\) is the length of the pipe.
Step 2: Find the beat frequency.
The pipes give \(40\) beats in \(10\,\text{s}\).
So, beat frequency is \[ f_b=\frac{40}{10} \] \[ f_b=4\,\text{Hz} \]
Step 3: Write frequencies of the two pipes.
For the first pipe, \[ L_1=50\,\text{cm}=0.50\,\text{m} \] So, \[ f_1=\frac{v}{2(0.50)} \] \[ f_1=v \] For the second pipe, \[ L_2=51\,\text{cm}=0.51\,\text{m} \] So, \[ f_2=\frac{v}{2(0.51)} \] \[ f_2=\frac{v}{1.02} \]
Step 4: Use the beat frequency formula.
Beat frequency is \[ f_b=|f_1-f_2| \] Thus, \[ 4=\left|v-\frac{v}{1.02}\right| \] \[ 4=v\left(1-\frac{1}{1.02}\right) \] \[ 4=v\left(\frac{1.02-1}{1.02}\right) \] \[ 4=v\left(\frac{0.02}{1.02}\right) \] \[ 4=v\left(\frac{2}{102}\right) \] \[ 4=v\left(\frac{1}{51}\right) \] \[ v=204\,\text{ms}^{-1} \]
Step 5: Final conclusion.
Hence, the speed of sound in this medium is \[ \boxed{204\,\text{ms}^{-1}} \]
For an open organ pipe, the fundamental frequency is \[ f=\frac{v}{2L} \] where \(v\) is the speed of sound and \(L\) is the length of the pipe.
Step 2: Find the beat frequency.
The pipes give \(40\) beats in \(10\,\text{s}\).
So, beat frequency is \[ f_b=\frac{40}{10} \] \[ f_b=4\,\text{Hz} \]
Step 3: Write frequencies of the two pipes.
For the first pipe, \[ L_1=50\,\text{cm}=0.50\,\text{m} \] So, \[ f_1=\frac{v}{2(0.50)} \] \[ f_1=v \] For the second pipe, \[ L_2=51\,\text{cm}=0.51\,\text{m} \] So, \[ f_2=\frac{v}{2(0.51)} \] \[ f_2=\frac{v}{1.02} \]
Step 4: Use the beat frequency formula.
Beat frequency is \[ f_b=|f_1-f_2| \] Thus, \[ 4=\left|v-\frac{v}{1.02}\right| \] \[ 4=v\left(1-\frac{1}{1.02}\right) \] \[ 4=v\left(\frac{1.02-1}{1.02}\right) \] \[ 4=v\left(\frac{0.02}{1.02}\right) \] \[ 4=v\left(\frac{2}{102}\right) \] \[ 4=v\left(\frac{1}{51}\right) \] \[ v=204\,\text{ms}^{-1} \]
Step 5: Final conclusion.
Hence, the speed of sound in this medium is \[ \boxed{204\,\text{ms}^{-1}} \]
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