# Maximum kinetic energy of a particle

I'm having difficulty solving the following SHM problem

A particle executes simple harmonic motion with amplitude $$A=20cm$$. At one instant it is at $$+\frac{1}{4}$$ the amplitude, moving away from equilibrium. At a time $$0.5$$ seconds later, the particle has $$\frac{1}{3}$$ the maximum speed, moving towards equilibrium, and has a positive acceleration. If the particle has a mass of $$2kg$$, find its maximum kinetic energy.

I know the maximum kinetic energy is $$E_{kmax}=\frac{1}{2}m\omega^2A^2$$, and tried solving for $$\omega$$ by setting up the equations

$$1)\space \vec x(t)=A\cos(\omega t+\phi)$$ $$2)\space \vec v(t)=-\omega A\sin(\omega t+\phi)$$

but didn't get anywhere. Any help?

You might as well let $$t=0$$ be the instant of the second sentence. That tells you that $$\cos (\phi)=\frac 14$$ and that $$-\omega \sin (\phi) \gt 0$$. That should give you enough information to compute $$\phi$$. Analyze the next sentence to find the value of $$0.5 \omega+\phi$$. That will give you $$\omega$$. Now you have everything you need to find the maximum kinetic energy from your expression.
• Will it be wrong to assume that $\phi=0$ for this problem? – Anson Pang Feb 14 at 23:34
• No, that will just change the time when $t=0$. I think my choice of origin is slightly easier because it lets you find $\phi$ from just one equation. If you set $\phi=0$ you will get a value for $\omega t$ from the first equation and $\omega(t+0.5)$ from the second, which works fine. – Ross Millikan Feb 14 at 23:40