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.

  • $\begingroup$ Will it be wrong to assume that $\phi=0$ for this problem? $\endgroup$ – Anson Pang Feb 14 at 23:34
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.