Question

A particle executing SHM oscillates between two fixed points separated by $20\text{cm}$. If its maximum velocity is $30\text{cm/s}$, find its velocity when the displacement is $5\text{cm}$ from the mean position.

• A. $15\sqrt{3}\text{cm/s}$
• B. $20\sqrt{3}\text{cm/s}$
• C. $10\sqrt{3}\text{cm/s}$
• D. $5\sqrt{3}\text{cm/s}$

Answer 1

The correct option is A $15\sqrt{3}\text{cm/s}$


Given displacement between two extreme positions is $20\text{cm}$
Amplitude $=\frac{20}{2}=10\text{cm}$
We know that $v=\omega \sqrt{A^2-x^2}$ and hence, the maximum velocity will be at mean position i.e. at $x=0$ which is $v_{max}=A\omega$

$\Rightarrow \frac{v_{max}}{v}=\frac{A}{\sqrt{A^2-x^2}}=\frac{10}{\sqrt{{10}^2-5^2}}=\frac{10}{5\sqrt{3}}=\frac{2}{\sqrt{3}}$
$\Rightarrow v_{max}=v.\frac{2}{\sqrt{3}}\Rightarrow v=\frac{30\times \sqrt{3}}{2}=15\sqrt{3}\text{cm/s}$