# Simple Harmonic Motion

1. May 8, 2010

### Lavace

1. The problem statement, all variables and given/known data
A spring with stiffness s is suspended vertically with a mass m attached to it's free end. When the mass comes to rest the spring is found to have been stretched by 0.0981m. The mass is then pulled down a further 0.1m, released from rest and found to execute SHM.

We're given the general form of the SHM equation, i.e x(t) = Acos(wt + psi).

i) Calculate numerical values for the following: The amplitute, constant phase angle, frequency, period, maximum and minimum speed of the mass, the maximum kinetic energy for a mass of 1kg.

2. Relevant equations
F =- kx
U = 1/2*kx^2

3. The attempt at a solution]

I don't particularly know where to start, or rather where I intended to start is incorrect.

I was thinking that we'd start off by calculating the spring constant (stiffness), k (called s in the script). But where this fits in to finding the amplitude I'm not sure.
Is the amplitude just 0.1m (as that is where the mass was released).

What is meant by constant phase angle? Psi?

To find the period, we will have to know omega(0), which is equal to 2PI/T, where T is the period. Omega(0) is equal to k/m, but we don't know the mass, we could substitute in Newton II (F=ma), but I'm not sure where this will lead.

Any pushes in the correct direction is a massive thank you!

2. May 8, 2010

### kuruman

Can you find an expression for the distance by which a spring of stiffness s is stretched when mass m is attached to it?

3. May 19, 2010

### Lavace

So here's my new solution:

F = mg - kd - kx , where d is the extension the string was stretched by.

Because mg = kd (from the equilibrium balance), F = -kx

We can then say ma = -kx, and hence a = -(k/m)x, where w = SQRT(k/m)

But then, why were we given the extension of when the mass was placed on the spring? Was this to calculate the stiffness for the latter questions?

4. May 19, 2010

### kuruman

How will you find the amplitude?