Wave Mechanics Vibration Of String Question 2
A 1 cm long string vibrates with fundamental frequency of 256 Hz. If the length is reduced to $ \frac{1}{4} $ cm keeping the tension unaltered, the new fundamental frequency will be
[BHU 1997]
Options:
64
256
512
D) 1024
Show Answer
Answer:
Correct Answer: D
Solution:
$ n\propto \frac{1}{l} $
Therefore $ \frac{n _{2}}{n _{1}}=\frac{l _{1}}{l _{2}} $
Therefore $ n _{2}=\frac{l _{1}}{l _{2}}n _{1}=\frac{1\times 256}{1/4}=1024Hz $
BETA
