# Spring Harmonic Motion

1. Jun 23, 2012

### SalsaOnMyTaco

Hi there, SalsaOnMyTaco here again.
1. The problem statement, all variables and given/known data

A 44 gram mass is attached to a massless spring and allowed to oscillate around an equilibrium according to:
y(t) = 1.2*sin( 3.1415*t ) where y is measured in meters and t in seconds

-What is the spring constant in N/m ?

HELP: Simple harmonic motion with the amplitude A is equivalent to the motion on a circle with the radius A, and the same angular frequency omega.
The force acting on an object with the mass m moving on a circle with the radius A with the angular frequency ω is
F_circ=m*A*ω2.
The force exerted by a spring with the constant k is equal to
F(x)=k*x, where x is the displacement.
Due to the analogy mentioned above, the two forces are equal at the point of maximum displacement (amplitude), that is,
F_circ=F(A).
You can solve this equation for omega in terms of k and m.
2. Relevant equations

T=2∏√(m/k)

3. The attempt at a solution
I have no idea how to approach this problem. Should I start from figuring out the Period to then solve for K on the above equation?

2. Jun 23, 2012

### SalsaOnMyTaco

Nvm, problem solved. Since i was given w=2pi/t, i solved for T and used T on the equation from OP to solve for K.