An object in SHM

The phase constant is an angle representing the starting point on of the object.

You have to find it by considering the initial conditions.

HINT: You know that the object is at position x(t)=0 when t=0. Can you use this information with the formulas you supplied to find the phase constant?

 

Hi bcjochim07,

I don't think either of those angles will give you the answer they are looking for. You started the particles at x=0 when t=0; now they want the first time t that the particle is at x=6cm.

You could also say that the arccosine gives an infinite number of answers to the calculation arccos(6/10)--just keep adding and subtracting (2 pi) to the angles. This corresponds to the fact that the oscillator (assuming it runs forever!) reaches 6cm an infinite number of times.

So to find the angle, you first find out which angle corresponds to the angle of the cosine function at t=0. That's the phase angle you found before, which was -pi/2 or the angle -1.578 radians.

So what is the first angle after -1.578 radians that is equivalent to arccos(6/10)? It is (5.35 radians - 2 pi), or about -0.933 radians.

If you plot out your cosine function, you can see that at t=0, the angle is -pi/2; at the first peak of your function it has the position of 10cm and the angle is 0 (that's why the cosine function is at a peak there), so the first time that the oscillator reaches 6cm must correspond to an angle between -pi/2 and 0.

 

if it starts at 0, just use

x=Asin(omega*time)

no phase shift needed.

 

bcjochim07,

There's two main ways to choose the trigonometric function for these problems. One way is to always choose the same function form such as:

[tex]
\begin{align}
x(t) &= A \cos(\omega t + \phi)\nonumber\\
v(t) &= - A \omega \sin(\omega t + \phi)\nonumber\\
a(t) &= - A \omega^2 \cos(\omega t + \phi)\nonumber
\end{align}
[/tex]

and then choose the phase constant phi so that the plot of [tex]\cos(\omega t + \phi)[/itex] matches the initial (t=0) position and velocity of the particular oscillator in your problem.

(In your problem here, note that [tex]\cos(\omega t - \pi/2)[/itex] looks like [tex]\sin(\omega t))[/itex], so at t=0 it starts at x=0 and is going up, which is what you want.)

The other way, if the oscillator is either at the amplitudes or the equilibrium point at t=0, is to choose a trig function so that the phase is automatically zero. These are:

[tex]
\begin{align}
x(t) = A \cos(\omega t) \Longleftarrow &\mbox{ at t=0 oscillator starts at positive amplitude}\nonumber\\
x(t) = -A \cos(\omega t) \Longleftarrow &\mbox{ at t=0 oscillator starts at negative amplitude}\nonumber\\
x(t) = A \sin(\omega t) \Longleftarrow &\mbox{ at t=0 oscillator is at x=0 and is moving upwards}\nonumber\\
x(t) = -A \sin(\omega t) \Longleftarrow &\mbox{ at t=0 oscillator is at x=0 and is movign downwards.}\nonumber
\end{align}
[/tex]

So if the oscillator at t=0 matches one of these positions, you can choose one of these forms and the phase will be zero. (If it's not clear why these choices match those four cases just plot out each x(t) and I think you'll see it.)

Of course if you change the form of x(t), then the forms of v(t) and a(t) will change also. You take the derivative of x(t) to find the v(t) for each special case, and the derivative of that for the a(t).

 

I think you made a calculation error. pi/2 is greater than the magnitude of 0.933 so you should get a positive time.

I don't think you can use 4.41 seconds. It is the same position, but that is the third time after t=0 that the particle gets there.

So at t=0, the particle is at x=0.

Solve pi/2*t -pi/2= -.933 to find the time that the particle is at x=6cm.

If you want to know how the other 2 angles you found fit in, follow the path of the particle at later times. At t=1 second (one-fourth of a period), the particle is at the postive amplitude.

If you then use the first angle you found in post #4 (.927= pi/2 *t -pi/2) that is the time that the particle has reached x=6cm the second time (and now moving downwards).

If you use the next angle you found in post #4, it is the time that the particle is back at x=6m after going all the way to the bottom point and then back up.