Question

A point particle of mass $0.1kg$ is executing S.H.M of amplitude of $0.1m$. When the particle passes through the mean position, its kinetic energi is $18\times {10}^{-3}J$. The equation of motion of this particle when the initial phase of oscillation is ${45}^o$ can be given by :

• A. $0.1\cos (6t+\frac{\pi }{4})$
• B. $0.2\sin (\frac{\pi }{4}+2t)$
• C. $0.4\sin (t+\frac{\pi }{4})$
• D. $0.1\sin (6t+\frac{\pi }{4})$

Answer 1

The correct option is D $0.1\sin (6t+\frac{\pi }{4})$

Kinetic Energy of particle is

$KE=\frac{1}{2}m{v}^2$

$v$ at mean position$=\omega A$

$8\times {10}^{-3}=\frac{1}{2}0.1{\omega }^2(o.1{)}^2$

${\omega }^2=16$

$\omega =4$

The equation is

$y=Asin(\omega t+\varphi )$

$y=0.1sin(4t+\frac{\pi }{4})$