Wave Mechanics Question 762
Question: Calculate the frequency of the second harmonic formed on a string of length 0.5 m and mass 2 ´ 10?4 kg when stretched with a tension of 20 N
[BHU (Med.) 2000]
Options:
A) 274.4 Hz
B) 744.2 Hz
C) 44.72 Hz
D) 447.2 Hz
Show Answer
Answer:
Correct Answer: D
Solution:
Mass per unit length $ m=\frac{2\times {{10}^{-4}}}{0.5}kg/m $
$ =4\times {{10}^{-4}}kg/m $ Frequency of 2nd harmonic $ n _{2}=2n _{1} $
$ =2\times \frac{1}{2l}\sqrt{\frac{T}{m}} $
$ =\frac{1}{0.5}\sqrt{\frac{20}{4\times {{10}^{-4}}}}=447.2Hz $
BETA
