A particle of mass 0.50 kg executes simple harmonic motion under force F = -50(N m⁻¹)x. The time period of oscillation is x/35 s. The value of x is _________. (Given π = 22/7) Numerical answer type. Work it out, then open the answer.

Answer: 22. F = -kx with k = 50 N/m. T = 2π√(m/k) = 2π√(0.5/50) = (2π)/(10) = π/5 T = (22)/(7× 5) = 22/35 s So x = 22.

A particle of mass 0.50 kg executes simple harmonic motion under… | JEE Main 2024 (9 Apr, Shift 2)

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Periodic Motion & SHM Basics

A particle of mass $0.50\text{ kg}$ executes simple harmonic motion under force $F = -50(\text{N m}^{-1})x$ . The time period of oscillation is $\frac{x}{35}\text{ s}$ . The value of $x$ is _________.

(Given $\pi = \frac{22}{7}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 22

$F = -kx$ with $k = 50\text{ N/m}$ .
$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.5}{50}} = \frac{2\pi}{10} = \frac{\pi}{5}$
$T = \frac{22}{7\times 5} = \frac{22}{35}\text{ s}$
So $x = 22$ .

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