Question

write the equation of motions of simple harmonic oscillator which has an amplitude of 5 cm and it executes 150 oscillation in 5 minutes with an initial phase of 45 degree also obtain the value of its maximum velocity

Answer 1

Task. Write the equation of motions of simple harmonic oscillator which has an amplitude of 5 cm and it executes 150 oscillation in 5 minutes with an initial phase of 45 degree also obtain the value of its maximum velocity.

Solution. The equation of a simple harmonic oscillation has the following form

$x(t)=A\cos(\omega t+\phi)$

where $A$ is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the phase.

Its velocity is then given by the formula:

$v(t)=x^{\prime}(t)=(A\cos(\omega t+\phi))^{\prime}=-A\omega\sin(\omega t+\phi).$

Hence the maximum of velocity is equal to

$\max v(t)=A\omega.$

In our case $A=5\ cm$, and $\phi=45{}^{\circ}=\frac{\pi}{4}$. Also, by assumption the oscillator executes 150 oscillation in 5 minutes. So its frequency is

$\nu=\frac{150}{5\ min}=\frac{150}{5*60\ sec}=\frac{150}{300\ sec}=0.5\ sec^{-1},$

whence the angular frequency is

$\omega=2\pi\nu=2\pi*0.5\ sec^{-1}=\pi\ rad/sec.$

Therefore the equation of motion for this oscillator is the following one:

$x(t)=5\cos(\pi t+\frac{\pi}{4})\ cm$

and the maximum of velocity is equal to

$\max v(t)=A\omega=5\ cm*\pi\ rad/sec=5\pi\ rad\,cm/sec.$

Answer. $x(t)=5\cos(\pi t+\frac{\pi}{4})\ cm,\quad\max v=A\omega=5\pi\ \ \mathrm{rad}\cdot\mathrm{cm/sec}.$