Calculating Force on an Oscillating Mass: A Simple Harmonic Motion Problem

[jumpingjack90]

Force on an oscillating mass?

Homework Statement

A 2.10 kg mass on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 6.00 N/m. The mass is displaced 3.09 m to the right from its equilibrium position and then released, which initiates simple harmonic motion. What is the force (including direction, choose the right to be positive) acting on the mass 4.18 s after it is released?

Homework Equations

F=kx
w=(sqrt)k/m
x(t)=xm*cos(w*t+phase costant)

The Attempt at a Solution

Find angular frequency
w=(sqrt)k/m=(sqrt)6.00/2.10=1.69rad/s
2-The displacement is given as a function of time:
x(t)=xm*cos(w*t+phase costant)
the function is:
x(t)=3.09*cos(1.69t)
the hooke law for spring is:
F=-kx
x(4.18)=3.09*cos(1.69*4.18)=3.06m
the force acting is:
F=-6.00*3.06=18.36N
Which is the incorrect answer. I'm not sure what I'm doing wrong. any help will be greatly appreciated. thanks!

 

[kuruman]

Did you forget the negative sign in front of 18.36 N ?

 

[RoyalCat]

Whoops, ignore whatever I wrote earlier, if you caught it, I misread your answer.
Everything looks peachy from here, negative sign aside.

For a simple harmonic oscillator:
$$x(t)=A\cos{(\omega t + \theta)}$$
Your initial conditions dictate that this be a cosine function with a phase constant 0 (Since its at its max amplitude at $$t=0$$)
$$x(t)=3.09\cos{(1.69t)}$$

Haha! I feel so dumb. I was operating under degrees instead of radians in my calculator. I suspect the thread-starter made the same mistake.

You're on the right track, but $$3.09\cos{(1.69*4.18)}$$ is approximately 2.195, and not 3.06.

 

[kuruman]

An additional issue is that

3.09*cos(1.69*4.18) does not give 3.06. You need to redo the calculation.

 

[jumpingjack90]

solved it! thanks for the input everyone!