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

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Topic Question Set

Q 1 :

A light hollow cube of side length 10 cm and mass 10 g, is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is $y π \times 10^{- 2} s$ , where the value of y is (Acceleration due to gravity, $g = 10 m / s^{2}$ , density of water $= 10^{3} kg / m^{3}$ ) [2025]

2
6
4
1

1

2

3

4

(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 $kg / m^{3}$

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

Q 2 :

Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is 100 g. The time period of the motion of the particle will be (approximately)

(take g = 10 ${ms}^{- 2}$ , radius of earth = 6400 km) [2023]

1 hour 24 minutes
1 hour 40 minutes
12 hours
24 hours

1

2

3

4

(1)

$T = 2 π \sqrt{\frac{R}{g}} = 2 π \sqrt{\frac{6400 \times 10^{3}}{10}}$

$= 2 π \times 8 \times 10^{2} = 16 π \times 10^{2} = 5024 s = 83.73 min$

$= 1 hr 23.73 min$

$\approx 1 hr 24 min$

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