Question

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 \( \dots \dots \dots \dots \). \[ \text{(Given } \pi = \frac{22}{7} \text{)} \]

Answer 1

Correct Answer: 22

The force is given by $F = -kx$, so $k = 50 \, \mathrm{Nm}^{-1}$. The mass $m = 0.50 \, \mathrm{kg}$. The time period for simple harmonic motion is: \[ T = 2\pi \sqrt{\frac{m}{k}}. \] Substituting $k = 50$ and $m = 0.5$: \[ T = 2\pi \sqrt{\frac{0.5}{50}} = 2\pi \sqrt{0.01} = 2\pi \cdot 0.1 = 0.2\pi \, \mathrm{s}. \] Given $T = \frac{x}{35}$, equating: \[ 0.2\pi = \frac{x}{35}. \] Substituting $\pi = \frac{22}{7}$: \[ 0.2 \cdot \frac{22}{7} = \frac{x}{35}. \] Simplifying: \[ x = 0.2 \cdot 22 \cdot 5 = 22. \]