Question

A particle is executing simple harmonic motion with an amplitude of $4cm$. At the mean position the velocity of the particle is $10cm/s$. The distance of the particle from the mean position when its speed becomes $5cm/s$ is

• A. $\sqrt{3}cm$
• B. $\sqrt{5}cm$
• C. $2\sqrt{3}cm$
• D. $2\sqrt{5}cm$

Answer 1

The correct option is C $2\sqrt{3}cm$

The velocity of teh particle executing SHM is given by:

$v_{max}=a\omega$

$\Rightarrow =\omega =\frac{v_{max}}{a}=\frac{10}{4}$

Now, $v=\omega \sqrt{a^2-y^2}$

$\Rightarrow v^2={\omega }^2(a^2-y^2)\Rightarrow y^2=a^2-\frac{v^2}{{\omega }^2}$

So, the distance of the particle will be:

$\Rightarrow y=\sqrt{a^2-\frac{v^2}{{\omega }^2}}=\sqrt{4^2-\frac{5^2}{(10/4{)}^2}}=2\sqrt{3}cm$