Question

A particle of mass m is performing simple harmonic motion along line AB with amplitude 2a with centre of oscillation as O. At time $t=0$ particle is at point $C(OC=a)$ and is moving towards B with velocity $v=a\sqrt{3}$ m/s. The equation of motion can be given by

• A. $x=2a(\sin t+\sqrt{3}\cos t)$
• B. $x=2a\sin (t+\frac{\pi }{6})$
• C. $x=a(\sin t+\sqrt{3}\cos t)$
• D. $x=a(\sqrt{3}\sin t+\cos t)$

Answer 1

Answer: (B), (D)

The correct options are
B $x=2a\sin (t+\frac{\pi }{6})$
D $x=a(\sqrt{3}\sin t+\cos t)$

Let the displacement equation of particle executing $S.H.M.$ is given by $x=Asin(wt+\varphi )$ ..........(1)

where $A$ is the maximum amplitude of oscillation

Thus velocity equation is $v=Awcos(wt+\varphi )$ ...............(2)

Given: $A=2a$

At time $t=0$, the particle is at $C$ which is $a$ distance away from the mean position moving away with velocity $v=a\sqrt{3}m/s$

Equation (1) becomes $x=2asin\left(\varphi \right)$ $(t=0)$

Thus $a=2asin\left(\varphi \right)⟹\varphi =\frac{\pi }{6}$ ............(4)

Also (4) satisfies the equation (2), $v=\left(2a\right)cos\frac{\pi }{6}=a\sqrt{3}$ (for $w=1$)

Hence the equation of motion of the particle is given by $x=2asin(t+\frac{\pi }{6})$

(for $w=1rad/s$)

Also by expanding, $x=2a[sintcos\frac{\pi }{6}+sin\frac{\pi }{6}cost]$

$⟹x=a(\sqrt{3}sint+cost)$