Oscillation and Simple Harmonic Motion | Physics JEE previous year questions with Solutions

(1)

Additional buoyant force,

$L^{2} x ρ g = m a_{net}$

$a_{net} = (\frac{L^{2} ρ g}{m}) x$

$ω^{2} = (\frac{L^{2} ρ g}{m}) \Rightarrow ω = \sqrt{(\frac{L^{2} ρ g}{m})}$

$T = 2 π \sqrt{\frac{m}{L^{2} ρ g}}$

where m = 10 g, L = 10 cm, $ρ$ = 1000 $kg / m^{3}$

$T = 2 π \sqrt{\frac{10 \times 10^{- 3}}{{(10 \times 10^{- 2})}^{2} \times 1000 \times 10}}$

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