Wave Mechanics Vibration Of String Question 25
Question: A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by a mass M, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of M is
[IIT-JEE (Screening) 2002]
Options:
A) 25 kg
B) 5 kg
C) 12.5 kg
D) 1/25 kg
Show Answer
Answer:
Correct Answer: A
Solution:
The frequency of vibration of a string $ n=\frac{p}{2l}\sqrt{\frac{T}{m}} $ Also number of loops = Number of antinodes.
Hence, with 5 antinodes and hanging mass of 9 kg.
We have p = 5 and T = 9g
Therefore $ n _{1}=\frac{5}{2l}\sqrt{\frac{9g}{m}} $ With 3 antinodes and hanging mass M We have p = 3 and T = Mg
Therefore $ n _{2}=\frac{3}{2l}\sqrt{\frac{Mg}{m}} $
$ \because $ n1 = n2
Therefore $ \frac{5}{2l}\sqrt{\frac{9g}{m}}=\frac{3}{2l}\sqrt{\frac{Mg}{m}} $
Therefore $ M=25 $ kg.
BETA
