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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?

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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.

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  • $\begingroup$ Will it be wrong to assume that $\phi=0$ for this problem? $\endgroup$ – Anson Pang Feb 14 at 23:34
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    $\begingroup$ 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. $\endgroup$ – Ross Millikan Feb 14 at 23:40

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