Wave Mechanics Question 595
Question: A small ball of density $ 4{\rho _{0}} $ is released from rest just below the surface of a liquid. The density of liquid increases with depth as $ \rho ={\rho _{0}}(1+ay) $ where $ a=2{{m}^{-1}} $ is a constant. Find the time period of its oscillation. (Neglect the viscosity effects).
Options:
A) $ \frac{2\pi }{\sqrt{5}}\sec $
B) $ \frac{\pi }{\sqrt{5}}\sec $
C) $ \frac{\pi }{2\sqrt{5}}\sec $
D) $ \frac{3\pi }{2\sqrt{5}}\sec $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] In equilibrium position $ mg=F _{B}=\frac{m}{4{\rho _{0}}}{\rho _{0}}(1+ay _{0})g\Rightarrow 4=1+ay _{0} $
$ y _{0}=\frac{3}{2}=1.5 $ Now displace it downward by $ \Delta y $
$ \
Therefore F=mg-\frac{m}{4{\rho _{0}}}{\rho _{0}}[1+a(y _{0}+\Delta y)]g $
$ =mg-\frac{mg}{4}[1+ay _{0}+a\Delta y] $
$ F=-\frac{mga}{4}\Delta y; $ Acceleration $ =-( \frac{ag}{4} )\Delta y $
$ \
Therefore T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{4}{ag}}=\frac{2\pi }{\sqrt{5}}\sec . $
BETA
