Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: An object in SHM

  1. Apr 19, 2008 #1
    1. The problem statement, all variables and given/known data
    An object in SHM oscillates with a period of 4 s and an amplitude of 10 cm. How long does the object take to move from x= 0 cm to x=6 cm?

    2. Relevant equations

    3. The attempt at a solution

    I think I should use the formulas

    v(t)= -omega*Asin(omega*t + phase constant)
    x(t) = Acos(omega*t + phase constant)

    But once again, I cannot find the phase constant and I do not understand the physical meaning of the phase constant. I really am having a lot of difficulty with these types of problems.
  2. jcsd
  3. Apr 19, 2008 #2


    User Avatar
    Homework Helper
    Gold Member

    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?
  4. Apr 19, 2008 #3
    so would the phase constant be -pi/2?
  5. Apr 19, 2008 #4
    then after that could I say 6cm=10cm*cos(pi/2 * t -pi/2) but here is the problem that I'm having... how do I know which angle to use since arc cos can give 2 values?

    the two values I get are .927= pi/2 *t -pi/2 and 5.35= pi/2*t -pi/2

    Is it 5.35 since it is moving to the right, and if I imagine a circle with an object rotating counterclockwise, it would be moving to the right at 5.35 rad?
  6. Apr 19, 2008 #5


    User Avatar
    Homework Helper

    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.
  7. Apr 20, 2008 #6
    if it starts at 0, just use


    no phase shift needed.
  8. Apr 20, 2008 #7
    so is the phase constant not -pi/2?
  9. Apr 20, 2008 #8
    I apparently don't understand the logic involved in these types of problems. Every time I think I might have figured it out, I find I don't really know what I'm doing.
  10. Apr 20, 2008 #9


    User Avatar
    Homework Helper


    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:

    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

    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:

    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

    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).
  11. Apr 20, 2008 #10
    Ok. That makes better sense now I think

    another question:You said: 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 I set the expression pi/2*t -pi/2= -.933 I will get a negative time. 5.35 radians is the really the same position as -.933 so could I say

    5.35= pi/2*t-pi/2 t=4.41s
  12. Apr 20, 2008 #11


    User Avatar
    Homework Helper

    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.
  13. Apr 20, 2008 #12
    Oh yes, plotting it out on my graphing calculator really helps me see it. Thanks!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook