wave_mechanics mock_test_waves_and_acoustics Question 9
Question: A string is under tension so that its length is increased by $ 1/n $ times its original length. The ratio of fundamental frequency of longitudinal vibrations and transverse vibrations will be
Options:
A) $ 1:n $
B) $ n^{2}:1 $
C) $ \sqrt{n}:1 $
D) $ n:1 $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Velocity of longitudinal waves$ v _{1}=\sqrt{\frac{Y}{\rho }} $ and velocity of transverse waves $ v _{2}=\sqrt{\frac{T}{m}}=\sqrt{\frac{T}{\rho s}} $
$ \therefore \frac{v _{1}}{v _{2}}=\sqrt{\frac{Y}{T/s}}=\sqrt{\frac{Y}{Y( \frac{\Delta l}{l} )}}=\sqrt{n} $
$ [ \therefore \Delta l=\frac{l}{n} ] $ Now $ f\propto v\therefore \frac{f _{1}}{f _{2}}=\frac{v _{1}}{v _{2}}=\sqrt{n} $ In the above expression, $ \rho $ = density of string, s = area of cross-section of string, Y= Young’s modulus.
BETA
