Simple Harmonic Motion of a hanging spring

[Winegar12]

Homework Statement

A 450 g object oscillates from a vertically hanging light spring once every .55 s. The object is released with the spring compressed by 10 cm from the equilibrium position. What is the velocity of the object 3.00 s after it is released?

Homework Equations

v=-[tex]\omega[/tex]Asin([tex]\omega[/tex]t+[tex]\phi[/tex]₀)
[tex]\Delta[/tex]L=mg/k

The Attempt at a Solution

Ok so I have found that [tex]\omega[/tex] is equal to 11.4. I tried to figure out k by putting it into the 2nd equation and found that it is equal to .075. I then subtracted that from .1 m (10 cm) which I thought was suppose to give me the A which I found to be .025. Obviously t is 3 s. I thought for [tex]\phi[/tex] it was equal to pi, but apparently I am wrong. Because when I put it into the equation to get velocity the answer is wrong. So I'm pretty sure that I have my angular frequency right, but I am either wrong on my amplitude (A) or on [tex]\phi[/tex]. But I'm not really sure. Can anyone help me out? Thanks!

 

[Vagn]

Your amplitude is 0.1, the amplitude is how far it is compressed at the starting point by conservation of energy. [itex] \phi [/itex] is 0 as your intial position is at the position where amplitude is max i.e. where [itex] \omega t + \phi = 0 [/itex] and then you should be able to get an answer.

Spoiler

I got -1.10 as an answer

 

[Winegar12]

I tried using 0 as well, but that answer isn't right either. I know the final answer is -.335 m/s...

 

[Vagn]

Amplitude is definitely 0, otherwise we need a new source of energy, as the mass is at rest at t=0, and [itex] \phi [/itex] must be similarly so for the same reason as above.

I've just noticed that I used cos instead of sin in my original answer, using sin I get an answer of -0.322m/s which is a bit better.

 

[rwisz]

Based on the given information, it seems that you have correctly calculated the angular frequency, \omega, and the spring constant, k. However, it appears that you have made a mistake in calculating the amplitude, A. The amplitude of a simple harmonic motion is equal to the maximum displacement from the equilibrium position, which in this case is 10 cm. Therefore, the amplitude should be 0.1 m, not 0.025 m. Additionally, the phase angle, \phi, is not equal to \pi in this case. The phase angle represents the initial position of the object and is dependent on the initial conditions of the system. Since the object is released from a compressed position, the phase angle should be 0. Therefore, the correct equation for the velocity of the object at t = 3 s would be:

v = -(11.4)(0.1)sin(11.4(3)+0) = -1.14 m/s

I would also suggest double checking your calculations and units to ensure accuracy.