- #1

zenterix

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- Homework Statement
- A particle performs harmonic oscillations along the ##x##-axis about the equilibrium position ##x=0##. The oscillation frequency is ##\omega=4\text{s}^{-1}##.

At a certain moment of time the particle has a coordinate ##x_0=25\text{cm}## and its velocity is equal to ##v_{x0}=100\text{cm/s}##.

Find the coordinate ##x## and the velocity ##v_x## of the particle ##t=2.40\text{s}## after that moment.

- Relevant Equations
- ##x(t)=a\cos{(\omega t-\phi)}##

##v(t)=-a\omega\sin{(\omega t-\phi)}##

##x(0)=a\cos{(-\phi)}=x_0##

##v(0)=-a\omega\sin{(-\phi)}=v_0##

##\implies \tan{(-\phi)}=-\frac{v_0}{\omega x_0}##

##\implies \phi=-\tan{\left (-\frac{v_0}{\omega x_0}\right )}##

The solution to this problem says that we can find ##a=\sqrt{x_0^2+(v_{x0}/\omega)^2}##

For the given values of ##\omega, x_0##, and ##v_{x0}## we have ##\phi=-\frac{\pi}{4}## and so we can find that

##\phi=-\frac{\pi}{4}##

##x(0)=a\frac{\sqrt{2}}{2}=x_0##

##a=\frac{2x_0}{\sqrt{2}}##

##x(0)=a\cos{(-\phi)}=x_0##

##v(0)=-a\omega\sin{(-\phi)}=v_0##

##\implies \tan{(-\phi)}=-\frac{v_0}{\omega x_0}##

##\implies \phi=-\tan{\left (-\frac{v_0}{\omega x_0}\right )}##

The solution to this problem says that we can find ##a=\sqrt{x_0^2+(v_{x0}/\omega)^2}##

**How do we find this expression?**For the given values of ##\omega, x_0##, and ##v_{x0}## we have ##\phi=-\frac{\pi}{4}## and so we can find that

##\phi=-\frac{\pi}{4}##

##x(0)=a\frac{\sqrt{2}}{2}=x_0##

##a=\frac{2x_0}{\sqrt{2}}##