A bob of mass 'm' suspended by a thread of length l undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density ${1 \over 4}$ times that of the bob and the length of the thread is increased by 1/3rd of the original length, then the time period of the simple harmonic oscillations will be :-
Solution
$T = 2\pi \sqrt {l/g}$<br><br>When bob is immersed in liquid<br><br>mg<sub>eff</sub> = mg $-$ Buoyant force<br><br>mg<sub>eff</sub> = mg $-$ v$\sigma$g ($\sigma$ = density of liquid)<br><br>$= mg - v{\rho \over 4}g$<br><br>$= mg - {{mg} \over 4} = {{3mg} \over 4}$<br><br>$\therefore$ ${g_{eff}} = {{3g} \over 4}$<br><br>${T_1} = 2\pi \sqrt {{{{l_1}} \over {{g_{eff}}}}}$<br><br>${l_1} = l + {l \over 3} = {{4l} \over 3},\,{l_{eff}} = {{3g} \over 4}$<br><br>By solving<br><br>${T_1} = {4 \over 3}2\pi \sqrt {l/g}$<br><br>${T_1} = {{4T} \over 3}$
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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