Position of Mass (Spring Mass Sytem)

[Kanashii]

Homework Statement

A force of 10 Newtons can stretch a spring by 0.04 m. Suppose a mass of 5 kg is attached to the lower end of the spring. We stretch the mass downward by 0.05 m from its equilibrium position and release it from rest. Determine the position of the mass relative to its equilibrium position at t = 0.5 seconds. Assume no damping.

Homework Equations

k = F/x
w= k/m
x(t) = A cos wt + B sin wt

The Attempt at a Solution

k = 10/ 0.04 = 250 N/m
x(0) = -0.05
x`(0) = 0
w= (250/5)^1/2 = 50^1/2
equation x(t) = -0.05cos( (50^1/2) t))
substituting t= 0.5, -0.049904837
but the answer is -0.04617 I do not know where I messed up.

 

[Jonathan Scott]

equation x(t) = -0.05cos( (50^1/2) t))
substituting t= 0.5, -0.049904837
but the answer is -0.04617 I do not know where I messed up.

You've taken the cosine of the angle in degrees, but it should be in radians.

 

[Kanashii]

You've taken the cosine of the angle in degrees, but it should be in radians.

Oooohhh
Thank you very much!