Question

A particle starts simple harmonic motion from the mean position. Its amplitude is a and total energy E. At one instant its kinetic energy is 3E/4. Its displacement at the instat is-

• A. $a/\sqrt{2}$
• B. $a/2$
• C. $\frac{a}{\sqrt{3/2}}$
• D. $a/\sqrt{3}$

Answer 1

The correct option is B $a/2$

The equation of the simple harmonic motion is $y=Asin\omega t$ with usual notations.

The total energy is $\left(\frac{1}{2}\right)m{\omega }^2{a}^2$ and the kinetic energy is $\left(\frac{1}{2}\right)m{\omega }^2\left(a^2-y^2\right).$

Therefore we have $\left(\frac{1}{2}\right)m{\omega }^2{a}^2=E$ and

$\left(\frac{1}{2}\right)m{\omega }^2\left(a^2-y^2\right)=\frac{3E}{4}.$

Dividing the second equation by the first , $1-\frac{y^2}{a^2}=\frac{3}{4}$ from which $y=\frac{a}{2}$.