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

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 $m = 0.5$:
$T = 2\pi \sqrt{\frac{0.5}{50}} = 2\pi \sqrt{0.01} = 2\pi \times 0.1 = 0.2\pi \, \mathrm{s}.$
Given $T = \frac{x}{35}$, equating:
$0.2\pi = \frac{x}{35}.$
Substituting $\pi = \frac{22}{7}$:
$0.2 \times \frac{22}{7} = \frac{x}{35}.$
Simplifying: $x = 0.2 \times 22 \times 5 = 22.$

Answer 2

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 $m = 0.5$:
$T = 2\pi \sqrt{\frac{0.5}{50}} = 2\pi \sqrt{0.01} = 2\pi \times 0.1 = 0.2\pi \, \mathrm{s}.$
Given $T = \frac{x}{35}$, equating:
$0.2\pi = \frac{x}{35}.$
Substituting $\pi = \frac{22}{7}$:
$0.2 \times \frac{22}{7} = \frac{x}{35}.$
Simplifying: $x = 0.2 \times 22 \times 5 = 22.$