Question

The maximum speed of a particle of mass executing SHM is ${\upsilon }_o$. What is the kinetic energy of the particle ' when it is at distance$\left(\frac{1}{\sqrt{2}}\right)$ times the amplitude form its mean position?

• A. $\frac{1}{2}{mv}_0^2$
• B. $\frac{1}{3}{mv}_0^2$
• C. $\frac{1}{4}{mv}_0^2$
• D. None of the above

Answer 1

The correct option is C $\frac{1}{4}{mv}_0^2$

Let equation of SHM is, $x=A\sin \omega t$

Then velocity of particle is given by, $v=\frac{dx}{dt}=A\omega \cos \omega t$

Maximum velocity, $v_o=A\omega$

Let at time t, particle is at distance $\frac{1}{\sqrt{2}}$ times the $A$

$x=\frac{A}{\sqrt{2}}=A\sin \omega t$

$\omega t=\frac{\pi }{4}$

At time t, velocity is, $v=A\omega \cos \frac{\pi }{4}=\frac{v_o}{\sqrt{2}}$

Kinetic energy $=\frac{1}{2}m{v}^2$

$K.E.=\frac{1}{2}m{\left(\frac{v_o}{\sqrt{2}}\right)}^2$

$K.E.=\frac{1}{4}m{v}_o^2$