Question

A particle executes SHM, with an amplitude of $10\text{cm}$. When the particle is at $6\text{cm}$ from the mean position, the magnitude of its velocity is equal to twice of its acceleration. Find the time period (in seconds)

• A. $4.1\text{s}$
• B. $9.42\text{s}$
• C. $1.17\text{s}$
• D. $14.8\text{s}$

Answer 1

The correct option is B $9.42\text{s}$

Given,
Amplitude $\left(A\right)=10\text{cm}$
Position of particle from mean position $\left(x\right)=6\text{cm}$
Since, particle executes SHM, we can use
$v=\omega \sqrt{A^2-x^2}$ ... (1)
$a={\omega }^{2}x$ ... (2)
$\therefore$ from (1) and (2) in $\left|v\right|=2\left|a\right|$
we get
$\omega \sqrt{A^2-x^2}=2{\omega }^{2}x$
$\Rightarrow \omega =\sqrt{\frac{A^2-x^2}{4x^2}}$
$\therefore$ Time period $\left(T\right)=\frac{2\pi }{\omega }$
$=2\pi \sqrt{\frac{4x^2}{A^2-x^2}}=\frac{2\times 3.14\times 2\times 6}{\sqrt{(10{)}^2-(6{)}^2}}$
$=\frac{2\times 3.14\times 2\times 6}{8}$
$=9.42\text{s}$