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jumpingjack90
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Force on an oscillating mass?
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?
F=kx
w=(sqrt)k/m
x(t)=xm*cos(w*t+phase costant)
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!
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!