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 1/4 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
(1) T
(2) \(\frac{3}{2}\)T
(3) \(\frac{3}{4}\)T
(4) \(\frac{4}{3}\)T
Correct answer: (4) \(\frac{4}{3}\)T
Explanation:
T = \(2 \pi \sqrt{\frac{\ell}{g}}\)
When bob is immersed in liquid
mgeff = mg – Buoyant force
mgeff = mg – vσg (σ = density of liquid)
= mg - v\(\frac{ρ}{4}\)g
= mg - \(\frac{mg}{4}\) = \(\frac{3mg}{4}\)
∴ geff = \(\frac{3g}{4}\)
T1 = 2π\(\sqrt{\frac{\ell_1}{g_{eff}}}\)
ℓ1 = ℓ + \(\frac{\ell}{3}\) = \(\frac{4 \ell}{3}\)
ℓeff = \(\frac{3g}{4}\)
By solving
T1 = \(\frac{4}{3}2 \pi \sqrt{\frac{\ell}{g}}\)
T1 = \(\frac{4}{3}\)T