Question

Two particles $p$ and $q$ describe simple harmonic motions of same period, same amplitude, along the same line about the same equilibrium position $O$. When $p$ and $q$ are on opposite sides of $O$ at the same distance from $O$ they have the same speed of $1.2m/s$ in the same direction, when their displacements are the same they have the same speed of $1.6m/s$ in opposite directions. The maximum velocity in $m/s$ of either particle is

• A. 2.8
• B. 2.5
• C. 2.4
• D. 2

Answer 1

The correct option is D 2

From the image,
$\varphi =-\theta +\omega t$.............................................$\left(1\right)$
$180-\varphi =\theta +\omega t$......................................$\left(2\right)$
substituting value of $\varphi$ from equn $\left(1\right)$ in $\left(2\right)$, we will get $\omega t=90$.
That means that particles will have to rotate by $9$0 in order to be again at same displacement.
Hence from $\left(1\right),$
we get $\varphi =-\theta +90$
now from figure,
$y=A\sin \theta$.....................................................$\left(3\right)$ and
$y^{\prime }=A\sin \varphi =A\sin (-\theta +90)=A\cos \theta =A\sqrt{1-{\sin }^2\theta }$
From eqn $\left(3\right)$
$y^{\prime }=\sqrt{A^2-y^2}$....................................................$\left(4\right)$
Initially magnitude of velocity, $=1.2m/s$
Therefore $1.2=\omega \sqrt{A^2-y^2}$......................$\left(5\right)$
and Finally $1.6=\omega \sqrt{A^2-y^{\prime 2}}$
that gives, $1.6=\omega y$ $($using eqn $\left(4\right))$ ......................$\left(6\right)$
Putting $\left(6\right)$ in eqn $\left(5\right)$ after doing a little algebra ,we will get , $\omega A=2m/s$, is the magnitude of maximum velocity which occurs at mean position.