PX155 - D3 - solving the SHM equation

• try: $x=A\cos \lambda t$
  $\dot{x}=-A\lambda \sin \lambda t$
  $\ddot{x}=-A{\lambda }^2\cos x=-{\lambda }^{2}x$
  if $\lambda =\omega$, we have a solution
  similarly, $x=B\sin \omega t$ is a solution
  • general solution:

- $\omega t$ plays the role of an angle in trig functions, so, $\omega$ acts like an angular velocity
- argument $\theta =\omega t⟹\omega =\frac{\theta }{t}=angularvelocity$
- solutions repeat every $\omega t=2\pi$ because $\cos (\omega t+2\pi )=\cos (\omega t)$, and $\sin (\omega t+2\pi )=\sin (\omega t)$
- $T=\frac{1}{f}=\frac{2\pi }{\omega }$
- alternate form of the solution:

$A^{\prime }=$ amplitude
$\varphi =$ phase angle
- comes from:
$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$
$A^{\prime }cos(\omega t+\varphi )=A^{\prime }cos(\varphi )cos(\omega t)-A^{\prime }sin(\varphi )sin(\omega t)$
-
- peaks at $\omega t+\varphi \pm 2n\pi =0$
- so, peak at $t=-\frac{\varphi }{\omega }\pm 2n\pi$

• eg: calculate the oscillation period when a mass of $0.1kg$ is attached to aspring with $k=100N{m}^{-1}$
  • $\omega =\sqrt{\frac{k}{m}},T=\frac{2\pi }{\omega }$
  • $T=2n/\sqrt{\frac{k}{m}}=0.199s$