Simple Harmonic Motion Example: An object at the end of a spring is released from x = 0 with initial velocity of v = -3 m/s at t=0. The spring constant k is 10 N/m. The mass of the object is 0.1 kg. What will be the frequency and period of oscillations? What equation describes this motion as a function of time?
Solution:
$$ w=\sqrt { \frac { k }{ m } } $$
$$ w=\sqrt { \frac { 10 }{ 0.1 } }=10 rad/s $$
$$ T=\frac { 2\pi }{ w } =0.628 s$$
$$ f=\frac { 1 }{ 0.628 } =1.6 Hz$$
If the cosine function is selected for the displacement $$x=A\cos { (wt+\varphi ) } $$
If the initial conditions for displacement are substituted,
$$0=A\cos { (10\cdot 0+\varphi ) } $$
$$0=A\cos { (\varphi ) }$$
For the above equation, A (Amplitude) can not be 0 since, a velocity of -3 m/s is given to mass spring system at equilibrium (x = 0) and the system is going to oscillate.
$$\cos { (\varphi ) } =0$$
$$\varphi =\frac { \pi }{ 2 } ,\frac { 3\pi }{ 2 }$$
If the initial conditions for velocity are substituted,
$$\dot { x } =v=-Aw\sin { (wt+\varphi ) } $$
$$\dot { x } =-3=-A10\sin { (10\cdot 0+\frac { \pi }{ 2 } ) }$$
$$A=0.3 m $$
or
$$\dot { x } =-3=-A10\sin { (10\cdot 0+\frac { 3\pi }{ 2 } ) }$$
$$A=-0.3 m $$
The cosine equations describe this motion as a function of time are
$$x=0.3\cos { (10t+\frac { \pi }{ 2 } ) } $$
$$x=-0.3\cos { (10t+\frac { 3\pi }{ 2 } ) } $$
If above procedure is solved for sine function of displacement, then sine equations describe this motion as a function of time are
$$x=-0.3\sin { (10t+0) } $$
$$x= 0.3\sin { (10t+\pi ) } $$