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Simple harmomic motion

  1. Mar 7, 2007 #1
    1. The problem statement, all variables and given/known data

    amplitude is kept at 4.0 cm and the vibration
    frequency increased to 50 Hz. T
    The particle passes through the origin at time t = 0.0 s, travelling in the +y-direction.

    (i) At what time does the valve first pass through the position y = 3.0 cm?
    (ii) What is the speed at this point?
    (iii) What is the acceleration at this point?
    (iv) At what positions will the valve have a speed of 6.0ms−1?

    2. Relevant equations

    w= 2*pi*f
    acceleration= -w^2*x
    v= rw

    3. The attempt at a solution

    w= 2*pi*f
    = 2*50*pi= 100pi

    acceleration= w^2*x
    = 100pi*4= 400pi

    dont know if this is correct?
    also dont know how to get the time and postions part (i and iv)
  2. jcsd
  3. Mar 7, 2007 #2
    What is the equation for the motion?

    What do you call the time derivative of the equation of motion?
    Last edited: Mar 7, 2007
  4. Mar 7, 2007 #3
    ad i) Do you know the equation for simple harmonic motion - i.e. the relation between position and time? Solving this part is just a matter of solving this equation for y = 3.

    ad iv) The same story. Just use the equation* for velocity of simple harmonic movement and solve for v = 6 m/s to get the time and use the equation in i) to get the position.

    *you might get this one by differentiating the equation in i)

    don't forget the minus sign - do you know what its significance is?
    Last edited: Mar 7, 2007
  5. Mar 14, 2007 #4

    equation of motion
    v2= u2+2as

    w= 2pi f
    w= 100 pi

    equation for SHM:x= asin wt
    3= 4 sin 100 pi t
    0.75= sin 100pi t
    inverse 0.75= 100pi t
    48.590...= 100 pi t
    t= 1.527 secs
  6. Mar 14, 2007 #5
    - indicates acceleartion towards the starting position
  7. Mar 14, 2007 #6
    v= rw
    = 3pi
  8. Mar 14, 2007 #7

    a= vw
    = 600pi

    is questions i- iv correct???
  9. Mar 14, 2007 #8


    User Avatar
    Gold Member

    i) is correct.
    ii) is wrong. To get a velocity differentiate the x=asinwt wrt t.
  10. Mar 14, 2007 #9
    ii)v= awcos wt

    a= 3
    w= 100pi
    using this values would this be ok ?
    and what about
  11. Mar 15, 2007 #10

    coz0= 1
    v= 300 pi

    is this correct?
  12. Mar 15, 2007 #11
    Ok for the first one , you need to form a equation and then solve it . It is given that the y - position is 0 at t=0. w=100pi and A=0.04m (always work with s.i units) . Using these info we can form the equation , y=A Sin(wt) , in this case it would be 0.03=0.04Sin((100pi)(t)) , solving for this will result in the answer!!!
  13. Mar 15, 2007 #12
    Sorry about the double post
    Last edited: Mar 15, 2007
  14. Mar 15, 2007 #13
    For part 2 , once again we need to form a equation, this time , we need to form the equation for velocity. Which would be v=-wACos(wt) (it is cos because at origin , y=0) From i) you have your t which you can sustitute into the equation and solve it!!
    Last edited: Mar 15, 2007
  15. Mar 15, 2007 #14
    Ok for part 3) , you need to form another equation. The equation would be a=-W^2ASin(wt). Using the ans from part 1) , you can sole the equation and get your ans!!!
  16. Mar 15, 2007 #15
    just for you info imy786 , v=Aw and a= Aw^2 give you the maximum velocity and maximum acceleration.
  17. Mar 15, 2007 #16


    i dont think it has a minus sign.

    book quotes

    v= Awcos wt
  18. Mar 15, 2007 #17

    a= Aw^2 sin (wt)
    v= Awcos wt

    you wrote these equations -

  19. Mar 15, 2007 #18
    URGENT NEED help............
  20. Mar 15, 2007 #19
    imy786 it depends on the situation of the question . Sketch a simple y vs t and v vs t graph side byside ,You will then see whether it is - or +
  21. Mar 15, 2007 #20
    I use cos instead of sin in my equations as my phase angle is 0 as it starts from y=0 at t=0. If u wish u can also use y= A Sin(wt + phi) then v= -AWCos(wt + phi) .Ooops my bad . Sorry .Like i said just draw a graph and you will see what to do.
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