Finding Spring Constant through Simple Harmonic Motion

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
SalsaOnMyTaco
Messages
32
Reaction score
0
Hi there, SalsaOnMyTaco here again.

Homework Statement



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.

Homework Equations



T=2∏√(m/k)

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?
 
Physics news on Phys.org
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.