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Homework Help: Time it takes an oscillating mass to reach a certain velocity

  1. Jan 31, 2008 #1
    Hi everybody, this problem is giving me trouble and I was wondering if you could give me any advice.

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

    The velocity of an object in simple harmonic motion is given by vx(t)= - (0.35 m/s)sin(20t + pi), where t is in s.

    What is the first time after t=0s at which the velocity is - 0.25 m/s?

    2. Relevant equations

    See the equation above.

    3. The attempt at a solution

    I simply plugged -.25 m/s in for vx(t) above and solved for t, so I got:

    -0.25 = -0.35sin(20t + pi)
    .714 = sin(20t + pi)

    When I solve for t I get t = -.397e-1 = -.0397s, but mastering physics (the program I use to do homework online) say it's wrong. I entered the answer as positive. Anybody have any idea? Thanks a lot!

    edit: I just went to see my professor and he helped me set it up, when finished we had sin(20t) = -5/7, but master physics says it's wrong.
    Last edited: Jan 31, 2008
  2. jcsd
  3. Jan 31, 2008 #2
    Trigonometric functions are periodic. This means that the function "repeats" itself over and over as you move along the X axis. If you visualize the sine involved in your problem, you'll realize that if you move sufficiently far to the right from -.0397 (and beyond 0), you'll get at a point p where vt(p) = vt(-.0397). That point is in fact -.0397 + Pi/10, because Pi/10 is the period of vt(x).
  4. Jan 31, 2008 #3
    Thanks for the reply. Does that mean that the time it reaches -.25m/s is .275? (-0.397 + pi/10). Where did that 10 come from? Thanks!
  5. Jan 31, 2008 #4
    The period of a sine function is given by 2pi/k, where k is the coefficient of x inside the bracket (i.e. for vt(x), the coefficient is 20). Really, it's not very useful for you to hear about this from me. You'd be better off learning better about trigonometric functions.
  6. Jan 31, 2008 #5
    I believe that I do have a decent understand of trigonometric functions, I just can't figure out why this problem is giving me so much trouble. I can't be doing much wrong...]

    I've got a graph of the wave drawn out and where the time will fall on the wave, but I just can't get it numerically.
    Last edited: Jan 31, 2008
  7. Jan 31, 2008 #6
    But you just did; .275 s.
  8. Jan 31, 2008 #7
    Thanks for the reply, but this darn program say it's wrong. I really don't like doing homework online.
  9. Jan 31, 2008 #8
    Did you calculate arcsine in radians?
  10. Jan 31, 2008 #9
    Yeah, arcsine(5/7) is .7956, then multiplying that my (1/20) as by my answer in the first post gives .0397
  11. Jan 31, 2008 #10
    You forgot to subtract pi before dividing by 20.

    Calculating it myself, I get around -0.12 for t. -0.12 + pi/10 = 0.2, approximately.
  12. Jan 31, 2008 #11
    Thanks a lot for the help, Werg. Any more input from anybody would be appreciated!
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