Question

The figure shows the variation of force acting on a particle of mass $400 \, g$ executing simple harmonic motion. The frequency of oscillation of the particle is

• A. $4s^{- 1}$
• B. $\left(5 / 2 \pi \right) \, s^{- 1}$
• C. $\left(1 / 8 \pi \right)s^{- 1}$
• D. $\left(1 / 2 \pi \right)s^{- 1}$

Answer 1

Answer: (B)

The slope of the curve is $\frac{F}{x}=-\frac{0.5}{5}=-0.1\frac{N}{c m}=-10 \, N/m$
But $F=-m\omega ^{2}x$ or $F/x=-m\omega ^{2}$
so, $-m\omega ^{2}=-10$ or $\omega ^{2}=\frac{10}{m}$
$\therefore \omega ^{2}=\frac{10}{4 \times 10^{- 1}}\Rightarrow \omega =\frac{10}{2}=5$
$\therefore f=\frac{\omega }{2 \pi }=\frac{5}{2 \pi } \, s^{- 1}$