# SIMPLE HARMONIC MOTION EXAMPLE

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

$$x=-0.3\sin { (10t+0) }$$

$$x= 0.3\sin { (10t+\pi ) }$$