Question

The system shown here is initially at rest. The spring is now compressed by $1\text{cm}$ and released. If acceleration of $6\text{kg}$ block, just after the spring is released, is $\frac{1}{3}{\text{cm/s}}^2$, time period of oscillation (in sec) of the system is $x\pi$. Find $x$.

Answer 1

Given:- ${\begin{array}{c}{\overrightarrow{F}}_{\text{net}}\end{array}|}_{m=6\text{kg}}=m{\overrightarrow{a}}_{\text{net}}$
$kx=m{a}_{\text{net}}$
$K\left[1\text{cm}\right]=6\left[\frac{1}{3}\frac{\text{cm}}{{\text{s}}^2}\right]$
$k\left[\frac{1}{100}\text{m}\right]=6\left[\frac{1}{300}\frac{\text{m}}{{\text{S}}^2}\right]$
$K=2\frac{\text{N}}{\text{m}}$
Now, from the concept of reduced mass :-
Reduced - Mass $\left(\mu \right)=\frac{m_1{m}_2}{m_1+m_2}=\frac{\left(6\right)\left(3\right)}{6+3}=2\text{kg}$
Now, Time period of oscillation for the system:-
$T=2\pi \sqrt{\frac{\mu }{k}}$
$T=2\pi \sqrt{\frac{2}{2}}=2\pi \text{sec}$
Hence, the value of $x$ is $2$.