Wave Mechanics Question 657
Question: A stretched wire 60 cm long is vibrating with its fundamental frequency of 256 Hz. If the length of the wire is decreased to 15 cm and the tension remains the same. Then the fundamental freuqency of the vibration of the wire will be
Options:
A) 1024
B) 572
C) 256
D) 64
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ L _{0}=60cm $
$ v _{0}=256Hz. $
$ v=\frac{1}{2L}\sqrt{\frac{T}{m}} $
$ \
Therefore v\propto \frac{1}{L} $
$ \frac{v _{1}}{v _{0}}=\frac{L _{0}}{L _{1}} $
$ \Rightarrow ,v _{1}=v _{0}\frac{L _{0}}{L _{1}}=256\times \frac{60}{15}=1024Hz. $
BETA
