Question

A particle is executing SHM with amplitude $A$, time period $T$, maximum acceleration $a_0$ and maximum velocity $v_0$. It starts from mean position at $t=0$ and at time $t$ it has the displacement $A/2$, acceleration $a$ and velocity $v$, then

• A. $t=\frac{T}{12}$
• B. $a=\frac{a_0}{2}$
• C. $v=\frac{v_0}{2}$
• D. $t=\frac{T}{8}$

Answer 1

Answer: (A), (B)

$t=\frac{T}{12}$
$a=\frac{a_0}{2}$

Let the displacement at t is $x=A\sin \omega t=A\sin \left(\frac{2\pi }{T}\right)t$
here,$x=\frac{A}{2}=A\sin \left(\frac{2\pi }{T}\right)t$
or, $\sin \left(\frac{2\pi }{T}\right)t=\frac{1}{2}=\sin \frac{\pi }{6}\Rightarrow t=\frac{T}{12}$
$\frac{dx}{dt}=\frac{2\pi }{T}A\cos \left(\frac{2\pi }{T}\right)t$
$\frac{d^{2}x}{d{t}^2}=-\left(\frac{2\pi }{T}{)}^{2}A\sin \right(\frac{2\pi }{T})t$
so, $v_{max}=v_0=\left(\frac{2\pi }{T}\right)A,a_{max}=a_0=(\frac{2\pi }{T}{)}^{2}A$
now at $t=T/12,$ the velocity,$v=\frac{dx}{dt}=\frac{2\pi }{T}A\cos \left(\frac{2\pi }{T}\right)\frac{T}{12}$
$=\frac{2\pi }{T}A\cos \frac{\pi }{6}=\frac{2\pi }{T}A\frac{\sqrt{3}}{2}$
$\therefore v=\sqrt{3}{v}_0$
at $t=T/12,$ the acceleration,$a=\frac{d^{2}x}{d{t}^2}=\left(\frac{2\pi }{T}{)}^{2}A\sin \right(\frac{2\pi }{T})\frac{T}{12}$
$=\frac{2\pi }{T}A\sin \frac{\pi }{6}=\frac{2\pi }{T}A\frac{1}{2}$
$\therefore a=\frac{a_0}{2}$