Finding time given positions in SHM

[jstevenson16]

Homework Statement

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is 0.090m , it takes the block 2.52s to travel from 0.090m to -0.090m . If the amplitude is doubled, to 0.180m , how long does it take the block to travel from 0.090m to -0.090m ?

Homework Equations

x=Acos(ωt+∅)

The Attempt at a Solution

Ive found the period to be 5.0s, angular frequency to be 1.26 rad/s, and ∅=1.05 rad. Rearranging above equation for time I got (cos-1(x/A)/ω)-∅=t, but this gives me disproportionally large answers. Any suggestions?

 

[gneill]

On what does the period of a spring-mass oscillator depend? (What's the usual formula for the period)? Do any of the relevant values change when the amplitude is changed?

 

[jstevenson16]

I am aware that the period does not depend on amplitude, however the problem asks for the time taken to cover half of the amplitude. Even with this knowledge I can't figure how to determine this time.

 

[gneill]

Okay. So while the period remains the same, the amplitude has changed. Call the new amplitude B = 0.180 m. Then a function describing position versus time for the oscillator is:

$x(t) = B cos(\omega t)$

You don't need to find a phase, just assume that the function describes the motion beginning at an amplitude maximum. Hence the simple cosine function.

Find the two times corresponding to x1 = 0.090 m and x2 = -0.090 m, and then take the difference.

 

[jstevenson16]

It worked! Thank you very much, it seems that I was thrown off by the phase and was ignoring that the times were dependent upon position and I needed to find the difference between the two.

 

[gneill]

It worked! Thank you very much, it seems that I was thrown off by the phase and was ignoring that the times were dependent upon position and I needed to find the difference between the two.

Excellent. Glad it worked out!