Question:
The mass and length of a string are 10 g and 100 cm respectively. If the tension in the string is increased from 400 N to 900 N, then the increase in the frequency of transverse vibration of the string is:
The mass and length of a string are 10 g and 100 cm respectively. If the tension in the string is increased from 400 N to 900 N, then the increase in the frequency of transverse vibration of the string is:
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Frequency is directly proportional to the square root of Tension.
Updated On: Jun 6, 2026
- 200 Hz
- 150 Hz
- 50 Hz
- 100 Hz
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The Correct Option is D
Solution and Explanation
Step 1: Concept
Frequency of a vibrating string: $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Step 2: Meaning
$f \propto \sqrt{T}$.
Step 3: Analysis
$\mu = 0.01$ kg / 1 m = 0.01 kg/m. $f_1 = \frac{1}{2(1)} \sqrt{\frac{400}{0.01}} = \frac{1}{2} \sqrt{40000} = \frac{200}{2} = 100$ Hz. $f_2 = \frac{1}{2(1)} \sqrt{\frac{900}{0.01}} = \frac{1}{2} \sqrt{90000} = \frac{300}{2} = 150$ Hz. Increase $= 150 - 100 = 50$ Hz. Wait, re-checking options. If I increase from 400 to 900, the ratio $\sqrt{T_2/T_1} = \sqrt{900/400} = 3/2 = 1.5$. $f_2 = 1.5 f_1$. Increase $= 0.5 f_1 = 50$ Hz. The option for 50 Hz is (C).
Step 4: Conclusion
The increase is 50 Hz.
Final Answer: (C)
Frequency of a vibrating string: $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Step 2: Meaning
$f \propto \sqrt{T}$.
Step 3: Analysis
$\mu = 0.01$ kg / 1 m = 0.01 kg/m. $f_1 = \frac{1}{2(1)} \sqrt{\frac{400}{0.01}} = \frac{1}{2} \sqrt{40000} = \frac{200}{2} = 100$ Hz. $f_2 = \frac{1}{2(1)} \sqrt{\frac{900}{0.01}} = \frac{1}{2} \sqrt{90000} = \frac{300}{2} = 150$ Hz. Increase $= 150 - 100 = 50$ Hz. Wait, re-checking options. If I increase from 400 to 900, the ratio $\sqrt{T_2/T_1} = \sqrt{900/400} = 3/2 = 1.5$. $f_2 = 1.5 f_1$. Increase $= 0.5 f_1 = 50$ Hz. The option for 50 Hz is (C).
Step 4: Conclusion
The increase is 50 Hz.
Final Answer: (C)
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