Question 91: Oscillations

Question

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 :-

Accepted Answer

$$T = 2\pi \sqrt {l/g} $$
When bob is immersed in liquid
mg
eff
= mg $$-$$ Buoyant force
mg
eff
= mg $$-$$ v$$\sigma$$g ($$\sigma$$ = density of liquid)
$$ = mg - v{ho  \over 4}g$$
$$ = mg - {{mg} \over 4} = {{3mg} \over 4}$$
$$	herefore$$ $${g_{eff}} = {{3g} \over 4}$$
$${T_1} = 2\pi \sqrt {{{{l_1}} \over {{g_{eff}}}}} $$
$${l_1} = l + {l \over 3} = {{4l} \over 3},\,{l_{eff}} = {{3g} \over 4}$$
By solving
$${T_1} = {4 \over 3}2\pi \sqrt {l/g} $$
$${T_1} = {{4T} \over 3}$$

Answer

T

Answer

$${3 \over 2}$$T

Answer

$${3 \over 4}$$T

Answer

$${4 \over 3}$$T

Oscillations

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