Question

A particle of mass $200g$ executes SHM. The restoring force is provided by the spring of spring constant $80N/m$. What is the magnitude of acceleration when mass is $2cm$ away from the equilibrium position?

• A. $8m/s^2$
• B. $10m/s^2$
• C. $12m/s^2$
• D. $20m/s^2$

Answer 1

The correct option is A

$8m/s^2$

Step. 1 Given Data,

$m=200g=200\times {10}^{-3}kg$

$K=80\raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$

$x=2cm=2\times {10}^{-2}m$

Step 2. Formula Used,

$\frac{d^2\overrightarrow{x}}{d{t}^2}=-{\omega }^2\overrightarrow{x}$,

Step 3. Calculating the acceleration

Putting all the given values in,

$$\begin{gathered} \frac{d^2\overrightarrow{x}}{d{t}^2}=-{\omega }^2\overrightarrow{x} \\ \frac{d^2\overrightarrow{x}}{d{t}^2}=-\left(\frac{k}{m}\right)\overrightarrow{x}(\frac{K}{m}={\omega }^2) \\ \frac{d^2\overrightarrow{x}}{d{t}^2}=-\left(\frac{80}{200\times {10}^{-3}}\right)2\times {10}^{-2}\left\{\right\} \end{gathered}$$

$K$ is the spring constant

$\omega$ is the angular momentum

$x$ is the magnitude of acceleration from the equilibrium position.

$m$ is the mass of a particle

$\frac{d^2\overrightarrow{x}}{d{t}^2}=8m/s^2$

Hence, the magnitude of the acceleration is $8m/s^2$ when the mass is $2cm$ away from the equilibrium position.

Hence option (A) is the correct answer.