Wave Mechanics Question 726
A stretched string of 1m length and mass $ 5\times {{10}^{-4}}kg $ is under tension of 20N. If it is plucked at 25cm from one end then it will vibrate with frequency
[RPET 1999; RPMT 2002]
Options:
A) 100 Hz
B) 200 Hz
C) 256 Hz
D) 400 Hz
Show Answer
Answer:
Correct Answer: B
Solution:
As we know plucking distance from one end $ =\frac{l}{2} $
Therefore $ 25=\frac{100}{2p} $
Therefore $ p=2 $ . Hence frequency of vibration $ n=\frac{p}{2l}\sqrt{\frac{T}{m}}=\frac{2}{2\times 1}\sqrt{\frac{20}{5\times {{10}^{-4}}}} $ = $ 100Hz. $
BETA
