Question

The time period of a particle in simple harmonic motion is $T$. Assume potential energy at mean position to be zero. After a time of $T/6$ it passes its mean position, its

• A. velocity will be half its maximum velocity
• B. displacement will be half its amplitude
• C. acceleration will be nearly $86\%$ of its maximum acceleration
• D. $KE=PE$

Answer 1

Answer: (A)

velocity will be half its maximum velocity

$y=a\sin \omega t=a\sin \frac{2\pi t}{T}$

$v=\frac{dy}{dt}=\omega a\cos \frac{2\pi t}{T}$

At $t=\frac{T}{6},v=\omega a\cos \left(\frac{2\pi }{T}\frac{T}{6}\right)=\frac{1}{2}\omega a$

or, $v=\left(v_{max}/2\right)$

$y=a\sin \frac{2\pi }{T}\times \frac{T}{6}=a\sin \frac{\pi }{3}$
It is not half of $a$.

Acceleration $=\frac{d^{2}y}{d{t}^2}=\frac{dv}{dt}={\omega }^{2}a\sin \frac{2\pi t}{T}$

$={\omega }^{2}a\sin \frac{\pi }{3}=0.86(AC{)}_{max}$

At this instant,
KE $=\frac{1}{2}m{v}^2=\frac{1}{2}m{\left(\frac{v_{max}}{2}\right)}^2=\frac{(KE{)}_{max}}{4}=\frac{1}{4}\left(TE\right)$

$\therefore PE=TE-KE=\frac{3}{4}\left(TE\right)$,

i.e, $KE\ne PE$

Hence Option A