Question

A particle of mass $200\text{g}$ executes simple harmonic motion. The restoring force is provided by a spring of spring constant $80{\text{Nm}}^{-1}$. What would be the time period?

• A. $0.93\text{s}$
• B. $0.62\text{s}$
• C. $0.31\text{s}$
• D. None of these

Answer 1

The correct option is C $0.31\text{s}$

Given, mass of particle $\left(m\right)=200\text{g}=0.2\text{kg}$
Spring constant $\left(K\right)=80\text{N/m}$
Time period of oscillation $\left(T\right)=2\pi \sqrt{\frac{m}{K}}$
$=2\times 3.14\times \sqrt{\frac{0.2}{80}}$
$=0.314\text{s}$