Wave Mechanics Question 589
Question: A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to $ T _{M} $ . If the Young’s modulus of the material of the wire is Y then $ \frac{1}{Y} $ is equal to: (g = gravitational acceleration)
Options:
A) $ [ 1-{{( \frac{T _{M}}{T} )}^{2}} ]\frac{A}{Mg} $
B) $ [ 1-{{( \frac{T}{T _{M}} )}^{2}} ]\frac{A}{Mg} $
C) $ [ {{( \frac{T _{M}}{T} )}^{2}}-1 ]\frac{A}{Mg} $
D) $ [ {{( \frac{T _{M}}{T} )}^{2}}-1 ]\frac{Mg}{A} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] As we know, time period, $ T=2\pi \sqrt{\frac{\ell }{g}} $ When additional mass M is added then $ T _{M}=2\pi \sqrt{\frac{\ell +\Delta \ell }{g}} $
$ {T _{\frac{M}{T}}}=\sqrt{\frac{\ell +\Delta \ell }{\ell }},or,{{( \frac{T _{M}}{T} )}^{2}}=\frac{\ell +\Delta \ell }{\ell } $ or, $ {{( \frac{T _{M}}{T} )}^{2}}=1+\frac{Mg}{Ay} $
$ [ \
Therefore ,\Delta \ell =\frac{Mg\ell }{Ay} ] $
$ \
Therefore \frac{1}{y}=[ {{( \frac{T _{M}}{T} )}^{2}}-1 ]\frac{A}{Mg} $
BETA
