Simple Harmonic Motion amplitude

[JKLS]

Homework Statement

An object in SHM oscillates with a period of 4.0 s and an amplitude of 10 cm. How long does the object take to move from x = 0.0 cm to x = 6.0 cm?

Homework Equations

T = 2*pi/w

x(t) = Acos(wt) or x(t) = Asin(wt)

The Attempt at a Solution

T = 4.0 s
A = 0.10 m

w = 2*pi/T
w = 2*pi/(4.0)
w = pi/2

What happens now?

 

[Delphi51]

Good start!
Ask yourself at what time is it at x=0? What time at x = 6 cm?
The difference between the two times is your final answer.

 

[JKLS]

I keep running into the issue of how to visualize the relative positions of these functions at different values. If I plug in 0.0 cm (or 0.0 m rather) into x(t) = Acos(wt), how do I know whether the function is moving positively ("up" a cosine crest) or negatively ("down" a cosine trough)? If it's positive, then the position difference will be the desired 6 cm. If it's negative, then the difference will be 10 + 10 + 6 = 26cm. How do I set this up appropriately in light of this?

 

[JKLS]

Good start!
Ask yourself at what time is it at x=0? What time at x = 6 cm?
The difference between the two times is your final answer.

I think what you're getting at is isolating t in each of x(t) = 0.00m and x(t) = 0.06m and finding the difference. However, this warrants a number (~23 seconds) that just doesn't make any sense given the period of 4 seconds.

 

[Delphi51]

I would use the x = 0.1*sin(πt/2).
Then at t = 0, x = 0.
At t = 1, x = 0.1 m or 10 cm.
So it will be at x = 6 cm sometime between 0 and 1 second. The motion is all in the same direction during this quarter of a period, so no up and down to worry about.

You could keep trying different times in that range until you get x = 0.06 or you could solve the equation for t and plug in x = .06.

 

[JKLS]

I would use the x = 0.1*sin(πt/2).
Then at t = 0, x = 0.
At t = 1, x = 0.1 m or 10 cm.
So it will be at x = 6 cm sometime between 0 and 1 second. The motion is all in the same direction during this quarter of a period, so no up and down to worry about.

You could keep trying different times in that range until you get x = 0.06 or you could solve the equation for t and plug in x = .06.

x = 0.1*sin(πt/2)
t = 2*asin(x/0.1)/π
t = 2*asin(0.06/0.1)/π
t = way too high.

Am I misinterpreting how to use this equation?

 

[Delphi51]

Ah, I see the problem. You were deceived by a little matter of units!
asin(x/0.1) = 36.9 degrees, which is 0.643 radians. t works out to about 0.4 seconds.
If you want to use degrees, then you must replace the π with 180 degrees in the formula.

 

[JKLS]

Ah, I see the problem. You were deceived by a little matter of units!
asin(x/0.1) = 36.9 degrees, which is 0.643 radians. t works out to about 0.4 seconds.
If you want to use degrees, then you must replace the π with 180 degrees in the formula.

UGH. Thank you. I'll try again.