Question

A particle executing simple harmonic motion has an angular frequency of 6.28 $s^{-1}$ and amplitude of 10 cm, find the speed at t = 1/6 s assuming that the motion starts from rest at t = 0.

• A. $54.4cm/s$
• B. $44.4cm/s$
• C. $64.4cm/s$
• D. $14.4cm/s$

Answer 1

The correct option is A $54.4cm/s$

At t = 0 the velocity is zero i.e. the particle is at an extreme The equation for displacement may be written as $x=A\cos \omega t$
The velocity is $v=-A\omega \sin \omega t$At $t=\frac{1}{6}s$ $v=-\left(0.1m\right)\left(6.28s^{-1}\right)\sin \left(\frac{6.28}{6}\right)$ $=(-0.628m/s)\sin \frac{\pi }{3}=54.4cm/s$