Wave Mechanics Vibration Of String Question 33
Question: Two wires are fixed in a sonometer. Their tensions are in the ratio 8 : 1. The lengths are in the ratio $ 36:35. $ The diameters are in the ratio 4 : 1. Densities of the materials are in the ratio 1 : 2. If the lower frequency in the setting is 360 Hz. the beat frequency when the two wires are sounded together is
[KCET 2003]
Options:
A) 5
B) 8
C) 6
D) 10
Show Answer
Answer:
Correct Answer: D
Solution:
Frequency in a stretched string is given by $ n=\frac{1}{2l}\sqrt{\frac{T}{\pi r^{2}\rho }}=\frac{1}{l}\sqrt{\frac{T}{\pi d^{2}\rho }} $ (d = Diameter of string)
Therefore $ \frac{n _{1}}{n _{2}}=\frac{l _{2}}{l _{1}}\sqrt{\frac{T _{1}}{T _{2}}\times {{( \frac{d _{2}}{d _{1}} )}^{2}}\times ( \frac{{\rho _{2}}}{{\rho _{1}}} )} $
$ =\frac{35}{36}\sqrt{\frac{8}{1}\times {{( \frac{1}{4} )}^{2}}\times \frac{2}{1}}=\frac{35}{36} $
$ \Rightarrow n _{2}=\frac{36}{35}\times 360=370 $ Hence beat frequency = $ n _{2}-n _{1}=10 $
BETA
