Question:
A vibrating string of certain length \({}'l'\) under a tension \({}'T'\) resonates with a mode corresponding to the first overtone of an air column of length \(75\ \text{cm}\) inside a tube closed at one end. The string also generates \(4\) beats/s, when vibrated along with a tuning fork of frequency \({}'n'\). Increasing the tension of the string reduces the beat frequency. The frequency \({}'n'\) of the tuning fork in Hz is \(\{\)Velocity of sound in air is \(340\ \text{m/s}\}\)
A vibrating string of certain length \({}'l'\) under a tension \({}'T'\) resonates with a mode corresponding to the first overtone of an air column of length \(75\ \text{cm}\) inside a tube closed at one end. The string also generates \(4\) beats/s, when vibrated along with a tuning fork of frequency \({}'n'\). Increasing the tension of the string reduces the beat frequency. The frequency \({}'n'\) of the tuning fork in Hz is \(\{\)Velocity of sound in air is \(340\ \text{m/s}\}\)
Show Hint
Find the string frequency from the first overtone of the closed pipe, then use the effect of raising tension on the beats to pick the fork frequency.
Updated On: Nov 12, 2026
- \(344\)
- \(336\)
- \(117.3\)
- \(109.3\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
A closed pipe resonates at odd harmonics. Its first overtone is the third harmonic: \(f=\frac{3v}{4L}\).
Step 2: Calculation
\(f=\frac{3\times 340}{4\times 0.75}=340\ \text{Hz}\). The string therefore vibrates at \(340\ \text{Hz}\). With 4 beats/s, the fork frequency is \(n=344\ \text{Hz}\) or \(336\ \text{Hz}\).
Step 3: Compare the options
Increasing the tension raises the string frequency \(\left(f\propto\sqrt{T}\right)\). The beat frequency falls only if the string frequency moves closer to the fork, so the fork must be above \(340\ \text{Hz}\). Hence \(n=344\ \text{Hz}\).
Final Answer:
The tuning fork frequency is \(344\ \text{Hz}\). \[ \boxed{\text{(A) }344} \]
A closed pipe resonates at odd harmonics. Its first overtone is the third harmonic: \(f=\frac{3v}{4L}\).
Step 2: Calculation
\(f=\frac{3\times 340}{4\times 0.75}=340\ \text{Hz}\). The string therefore vibrates at \(340\ \text{Hz}\). With 4 beats/s, the fork frequency is \(n=344\ \text{Hz}\) or \(336\ \text{Hz}\).
Step 3: Compare the options
Increasing the tension raises the string frequency \(\left(f\propto\sqrt{T}\right)\). The beat frequency falls only if the string frequency moves closer to the fork, so the fork must be above \(340\ \text{Hz}\). Hence \(n=344\ \text{Hz}\).
Final Answer:
The tuning fork frequency is \(344\ \text{Hz}\). \[ \boxed{\text{(A) }344} \]
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