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 energy is $8\times {10}^{-3}Joule$. Obtain the equation of motion of this particle if this initial phase of oscillation is ${45}^{\circ }$.

• A. $y=0.1\sin (\pm 4t+\frac{\pi }{4})$
• B. $y=0.2\sin (\pm 4t+\frac{\pi }{4})$
• C. $y=0.1\sin (\pm 2t+\frac{\pi }{4})$
• D. $y=0.2\sin (\pm 2t+\frac{\pi }{4})$

Answer 1

The correct option is A $y=0.1\sin (\pm 4t+\frac{\pi }{4})$

The displacement of a particle in S.H.M. is given by:
$y=a\sin (\omega t+\varphi )$
$velocity=\frac{dy}{dt}=\omega a\cos (\omega t+\varphi )$
The velocity is maximum when the particle passes through the mean position i.e.
${\left(\frac{dy}{dt}\right)}_{max}=\omega a$
The kinetic energy at this instant is given by
$\frac{1}{2}m{\left(\frac{dy}{dt}\right)}_{max}^2=\frac{1}{2}m{\omega }^2{a}^2=8\times {10}^{-3}joule$
or $\frac{1}{2}\times \left(0.1\right){\omega }^2\times (0.1{)}^2=8\times {10}^{-3}$
Solving we get $\omega =\pm 4$
Substituting the values of $a,\omega$ and $\varphi$ in the equation of S.H.M., we get
$y=0.1\sin (\pm t+\pi /4)metre$.