# Simple harmonic motion

1. Oct 22, 2009

### vipertongn

1. The problem statement, all variables and given/known data

A moving particle displays simple harmonic motion with an amplitude of 3 cm. What position will it be for it to have half of its maximum speed?

2. Relevant equations

vmax= omega*A

3. The attempt at a solution

I divided the amplitude by 2 but I believe that's an incorrect method. was i suppose to deriviate x(t)=omega*A*sin(omega*t+phi)

2. Oct 23, 2009

### Proofrific

I'm going to assume that the particle acts like a mass on a spring. Then, we can solve this problem using conservation of energy.

At the system's biggest displacement (3cm), there is no kinetic energy, so the total energy of the system is just the energy from the spring:

$$E_{tot} = \frac{1}{2} k A^2$$​

At the equilibrium point, there is no spring energy. All of the energy is kinetic. Then, we can find the maximum velocity:

$$E_{tot} = \frac{1}{2} m v_{max}^2$$

$$\frac{1}{2} k A^2 = \frac{1}{2} m v_{max}^2$$

$$\sqrt{\frac{k}{m}} A = v_{max}$$​

Note that $$\sqrt{k/m}$$ is usually called $$\omega$$, so this is simply the equation you gave us in your post.

Now, we want to find the position of the mass when the velocity is half max. At that point, there is both spring energy and kinetic energy.

$$E_{tot} = \frac{1}{2} k x^2 + \frac{1}{2} m \left( \frac{v_{max}}{2} \right)^2$$

$$\frac{1}{2} k A^2 = \frac{1}{2} k x^2 + \frac{1}{2 \cdot 4} m \frac{k}{m} A^2$$

$$A^2 = x^2 + \frac{1}{4} A^2$$

$$\frac{3}{4} A^2 = x^2$$

$$\sqrt{\frac{3}{4}} A = x$$​

So, the position of the mass when it's traveling at half the maximum speed is $$x = \sqrt{\frac{3}{4}} A$$.