Question

A string under tension of 129.6 N produces 10 beats/second when it vibrates along with a tuning fork. When the tension in the string is increased to 160 N, it vibrates in unison with the tuning fork. Then frequency of the tuning fork is

• A. 100 Hz
• B. 110 Hz
• C. 90 Hz
• D. 220 Hz
• E. 95 Hz

Hint:

Frequency of string depends on square root of tension.

Answer 1

The Correct Option is A

Concept: Frequency of a stretched string: \[ f \propto \sqrt{T} \] Beat frequency: \[ f_{beat} = |f_1 - f_2| \]

Step 1: Let frequency of tuning fork = \(f\), string frequency at 129.6 N = \(f_1\). \[ |f_1 - f| = 10 \]

Step 2: At tension 160 N, string is in unison. \[ f_2 = f \]

Step 3: Use relation \(f \propto \sqrt{T}\). \[ \frac{f_1}{f_2} = \sqrt{\frac{129.6}{160}} \]

Step 4: Simplify ratio. \[ \frac{f_1}{f} = \sqrt{0.81} = 0.9 \Rightarrow f_1 = 0.9f \]

Step 5: Use beat condition. \[ |0.9f - f| = 10 \Rightarrow 0.1f = 10 \]

Step 6: Solve. \[ f = 100 \, \text{Hz} \]

Step 7: Conclusion. \[ \boxed{100 \, \text{Hz}} \]