Calculating Time for Simple Harmonic Motion - \pi

[rhodium]

$$\pi$$Hi everyone,

I hope you can help me out with the following question. I am mainly confused about the sign

Homework Statement

A block in SHM with T= 4.0 s and A=0.1 m. How long does it take the object to move from x=0 m to x= 0.06 m.

Homework Equations

x=Acos(wt + $$\phi$$)
w=2pi/T

where T is period, w is angular frequency, phi is phase angle and A is amplitude.

The Attempt at a Solution

Using eq2, i solved for w, which is $$\pi$$/2.
Then I set eq1 equal to 0. The value of phi is thus + or - $$\pi$$/2. Since the object is assumed to be moving to the right (as it would have to if it wants to go from 0 to 0.06 m),. then we take the negative phase angle. Then, back to eq1, I put

0.06=0.1cos(($$\pi$$/2)t - $$\pi$$/2)

NOW, this is were my problem is.
Apparently, there are then two possible answers,

either 0.927 = ($$\pi$$/2)t - $$\pi$$/2, which gives 1.59 s
or - 0.927 = ($$\pi$$/2)t - $$\pi$$/2. which gives 0.41 s

Firstly, I don't understand when are we supposed to have a + and - option. Secondly, I don't trust the answer given, which is 0.41 s. Please help.

 

[Doc Al]

Homework Equations

x=Acos(wt + $$\phi$$)
w=2pi/T

Note that you can also use:
x=Asin(wt + $$\phi$$)
Which would eliminate the annoying phase factor, since x = 0 at t = 0:
x=Asin(wt)

0.06=0.1cos(($$\pi$$/2)t - $$\pi$$/2)

NOW, this is were my problem is.
Apparently, there are then two possible answers,

either 0.927 = ($$\pi$$/2)t - $$\pi$$/2, which gives 1.59 s
or - 0.927 = ($$\pi$$/2)t - $$\pi$$/2. which gives 0.41 s

Realize that you want the first solution after t = 0. Since t = 0 corresponds to θ = -π/2 = -1.57 radians, θ = -0.927 radians is the first solution. The +0.927 solution corresponds to the block being at x = 0.06 but going to the left. (Since the block goes back and forth, there are an infinite number of times that it's at x = 0.06; you want the first time it reaches that point.)

 

[thomate1]

Hello there,

It seems like you have made some good progress in solving this problem. To clarify, the reason there is a + and - option for the phase angle is because cosine is an even function, meaning that it is symmetric about the y-axis. This means that there are two possible values for the phase angle that will give the same x-value at a given time. So, in this case, both +pi/2 and -pi/2 will result in an x-value of 0.06 m at a certain time.

As for the given answers, it is important to remember that the equation x=Acos(wt + phi) describes the position of the object at any given time, not just when it is moving from x=0 to x=0.06. So, when we solve for t, we need to consider all possible values of phi that will give an x-value of 0.06. In this case, both +pi/2 and -pi/2 will give 0.06, so both answers are correct. However, it is important to note that the negative phase angle will give the time at which the object is moving in the positive direction (from x=0 to x=0.06), while the positive phase angle will give the time at which the object is moving in the negative direction (from x=0.06 to x=0).

I hope this helps clarify the problem for you. Keep up the good work in solving SHM problems!