Energy in an Oscillating Spring

[gmmstr827]

Homework Statement

A mass of 240 g oscillates on a horizontal frictionless surface at a frequency of 3.0 Hz and with amplitude of 4.5 cm.
a) What is the effective spring constant for this motion?
b) How much energy is involved in this motion

m = 240 g = 0.24 kg
A = 4.5 cm = 0.045 m
f = 3.0 Hz
g = 9.8 m/s^2

Homework Equations

k= F/x = mg/x
E = K + U
K = 1/2*m*v^2
U = 1/2*k*x^2
v = ±ε√(A^2-x^2)
ε = 2πf = √(k/m)
v=0 when x=±A since the object stops briefly before traveling back towards the equilibrium.
^^^ That means K=0

The Attempt at a Solution

a) k= F/x = mg/x = [(0.240 kg)(9.8 m/s^2)]/(0.045 m) ≈ 52.27 N/m

b) E = 1/2*(52.27 N/m)(.045m)^2 = 0.053 J

^^^ Does this all look correct?

 

[Delphi51]

a) k= F/x = mg/x

Why this formula? We aren't given the force of the spring and it will not be related to gravity in this situation - "on a horizontal surface". Use a different spring formula that makes use of one of the given quantities.

b) should be okay once you have the correct value for k.

 

[SammyS]

Homework Statement

A mass of 240 g oscillates on a horizontal frictionless surface at a frequency of 3.0 Hz and with amplitude of 4.5 cm.
a) What is the effective spring constant for this motion?
b) How much energy is involved in this motion

m = 240 g = 0.24 kg
A = 4.5 cm = 0.045 m
f = 3.0 Hz
g = 9.8 m/s^2

Homework Equations

k= F/x = mg/x
E = K + U
K = 1/2*m*v^2
U = 1/2*k*x^2
v = ±ε√(A^2-x^2)
ε = 2πf = √(k/m)
v=0 when x=±A since the object stops briefly before traveling back towards the equilibrium.
^^^ That means K=0

The Attempt at a Solution

a) k= F/x = mg/x = [(0.240 kg)(9.8 m/s^2)]/(0.045 m) ≈ 52.27 N/m

b) E = 1/2*(52.27 N/m)(.045m)^2 = 0.053 J

^^^ Does this all look correct?

Solve : 2πf = √(k/m) for k.

 

[gmmstr827]

Thanks for the tips!

Ok, so solving for k in 2πf = √(k/m) gets me k = 4m(πf)^2.
So, k = 4*(.240 kg)*(3.0 Hz)^2*(π)^2 ≈ 85.27 N/m

For part b, using the new k value gets me an answer of approximately 0.086 J. Is that correct?