A body of mass $1 \mathrm{kg}$ is executing simple harmonic motion (SHM). Its displacement $y$ (in $\mathrm{cm}$) at time t given by $y=\left[6\sin \left(100\mathrm{t}+\frac{\pi }{4}\right)\right]\mathrm{cm}.$ Its maximum kinetic energy is

Given that the mass of the particle, $m=1 \mathrm{kg}$ and displacement, $y=6\sin (100t+\pi /4)$

The general equation of $\mathrm{SHM}$ is given by

$y=A\sin \left(\omega t+\varphi \right)$ 

Where $A, \omega \text{and} \varphi$ are amplitude, angular frequency and phase difference respectively. 

$\Rightarrow A=6 \mathrm{cm}$, $\omega =100 {\mathrm{s}}^{-1}$

Now, the velocity of the particle is the rate of change of linear velocity. It means,

$v=\frac{\mathrm{d}y}{\mathrm{d}t}=A\omega \cos \omega t=\omega \sqrt{A^2-y^2}$ 

Kinetic energy becomes,

$\mathrm{K}.\mathrm{E}.=\frac{1}{2}m{v}^2=\frac{1}{2}m{\left(\omega \sqrt{A^2-y^2}\right)}^2=\frac{1}{2}m\left({\omega }^2\left(A^2-y^2\right)\right)$

For maximum kinetic energy at $y=0$, 

$$\begin{gathered} \mathrm{K}.\mathrm{E}.=\frac{1}{2}m\left({\omega }^2\left(A^2-0^2\right)\right)=\frac{1}{2}m\left({\omega }^2\left(A^2\right)\right) \\ \Rightarrow \mathrm{K}.\mathrm{E}.=\frac{1}{2}\times 1\times {\left(100\right)}^2\times {\left(\frac{6}{100}\right)}^2=18 \mathrm{J} \end{gathered}$$