Wave Mechanics Vibration Of String Question 52
Question: In an experiment with sonometer a tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. Tension of the string remaining constant the frequency of the second tuning fork is
[KCET 2005]
Options:
A) 163.84 Hz
B) 400 Hz
C) 320 Hz
D) 204.8 Hz
Show Answer
Answer:
Correct Answer: D
Solution:
In case of sonometer frequency is given by $ n=\frac{p}{2l}\sqrt{\frac{T}{m}} $
Therefore $ \frac{n _{2}}{n _{1}}=\frac{l _{1}}{l _{2}}\Rightarrow n _{2}=\frac{25}{16}\times 256=400Hz $
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