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

General SHM Question

  1. Dec 18, 2006 #1
    Here's the actual question that started my wondering about the correct representation of a shm function:

    1. The problem statement, all variables and given/known data

    A 0.500 kg mass attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.0 cm. Calculate (a) the maximum value of its speed and acceleration, (b) the speed and acceleration when the mass is 6.00 cm from the equilibrium position, and (c) the time it takes the mass to move from x = 0 to x = 8.00 cm.

    2. Relevant equations

    [tex]\omega = \sqrt{ k / m }[/tex]
    [tex]x = A \cos(\omega t + \phi)[/tex]

    3. The attempt at a solution

    Certainly, (a) is simple, and in it we determine [tex]\omega = \sqrt{8/.5} = 4[/tex], but for (b), why would we not use [tex]x = 10 \cos (4 t )[/tex]?? The solution's manual uses [tex]x = 10 \sin(4 t)[/tex]. Why would I use sine in this case and how do I tell the difference?

  2. jcsd
  3. Dec 18, 2006 #2
    It's the same to use sine or cosine, for calculating the speed and the acceleration, as the exercise ask you to do. To see the difference give values to t and compare the elongation in both cases. For exemple, for t=0, with the sine function you got x=0, but with the cosine function you have x=10 ?! Do you see the difference?
  4. Dec 18, 2006 #3
    yes I see the difference but in this specific case, is there anything in the question that would cause me to want to use the sine function and not cosine?

    For instance, if, starting from rest, the spring is stretched, we know to use the sine function, since x = A sin(omega t) ; if t = 0, then x = 0, which doesn't follow the description; hence we use x = A cos (omega t).

    But for this specific case, there doesn't seem to be some kind of indicator saying, "Use the sine function" so it seems like an ambiguous question unless, of course, there is something I'm not seeing.

    Do I understand what you mean sufficiently?


    The real issue involces part (b); consider [tex] x = A \cos(\omega t)[/tex] and [tex]x = A \sin(\omega t)[/tex]; we need to solve for t:

    In the first case, [tex]t = \frac{\cos^{-1}(x/A)}{\omega}[/tex] and in the other: [tex]t = \frac{ \sin^{-1}(x/A) }{\omega}[/tex]

    When using the cosine function, the answer comes up as t = .232, as where in the second case, t = .161, from which come the correct values in the back of the book. How was I supposed to know this?
    Last edited: Dec 18, 2006
  5. Dec 18, 2006 #4
    You can choose to use the sine or the cosine function, the difference is just at the phase angle.

    Why are you calculating time for part (b)? It's more complicated to go that way. Conservation of mechanical energy gives you speed. Hooke's Law gives you acceleration.
  6. Dec 18, 2006 #5
    yes, i guess so. i'm not quite there yet! thanks!
  7. Dec 18, 2006 #6

    Doc Al

    User Avatar

    Staff: Mentor

    I assume you mean part (c), where you are asked to calculate the time. Careful--you really want the change in time. What time does each function assign to the point x = 0? (That's where the phase difference comes in--each function uses a different starting point for describing the motion.)
  8. Dec 18, 2006 #7

    Doc Al

    User Avatar

    Staff: Mentor

    On second thought, you really did mean part (b). No problem, a similar comment applies. The times are different because they measure time from a different starting point. But that shouldn't matter. Pick sine or cosine and stick to it--you'll get the same answer each way.
  9. Dec 18, 2006 #8
    Thanks for correcting my stupid help. I am not being ironic.
  10. Dec 18, 2006 #9

    Doc Al

    User Avatar

    Staff: Mentor

    I don't think your help was stupid at all--I was not correcting you, I was trying to help lizzyb! :smile:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook