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Homework Help: SHM question with no other details

  1. Aug 18, 2008 #1
    Qns: A block rest on a flat plate that executes vertical simple harmonic motion with a period of 1.5s. What is the maximum amplitude of the motion if the block were to be in contact with the plate thoughout the motion?

    Can help me how to start?...there is not much details in this questions.
  2. jcsd
  3. Aug 18, 2008 #2
    You really don't need anymore detail than what the question gives. You need to find out how fast can the flat plate travel as not to give the block enough momentum to leave the plate when it starts to travel in the negative direction. Since you know the period is 1.5s, the velocity of the plate is then a function of its amplitude or displacement. The only assumption to be made is that the plates position is sinusoidal and not stepped.

    Hint: When the plate reaches its top peak, the block should experience an acceleration of 0.
    Last edited: Aug 18, 2008
  4. Aug 18, 2008 #3
    still catch no ball... :(
  5. Aug 18, 2008 #4
    Ok, find a ball or box or some solid object. Put it in your open hand and start moving your hand up and down slowly. Then start moving it up and down faster until the ball in your hand starts to leave it when you move your hand from going up to going down. This the point you need to calculate by finding the distance you move your hand in a certain amount of time. You know the acceleration of gravity(9.81 m/s^2) and now you need to find the acceleration of the plate and ball. You will find when you start manipulating the equations that the mass of the ball cancels out.
  6. Aug 19, 2008 #5
    man, i understand all of it...but getting the equations up is mind boggling. These are what i gotten.

    velocity= 2A/1.5s


    at the top peak which can be either -A or A
    acceleation = kA=0

    therefor A=k at the top peak.

    anybody can give the equations to play around with???
  7. Aug 19, 2008 #6


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    Hi hemetite,

    The acceleration is not zero at the amplitude.

    In this case you're looking for the amplitude at which the block is just barely leaving the plate at the highest point. The reason it is leaving the plate is because the spring is pulling the plate down faster than the ball can keep up. So if the block has lost contact with the plate, what is its acceleration?

    Once you have that, how is maximum acceleration related to amplitude?
  8. Aug 19, 2008 #7
    okay.... Acceleration of the plate cannot be more than 9.81ms or else the box will lose contact.

    diSplacement is proportional to force. So at the peak.

    m9.81=A cos(wt + teta)

    9.81= -w2 A cos(wt + teta)


    Correct so far?
  9. Aug 19, 2008 #8


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    That's right; so you want to find the amplitude that gives a maximum acceleration of 9.81 m/s.

    This does not look right; you have a force on the left and distance on the right.

    These last two are right and will give you the answer. What do you get for A?
  10. Aug 20, 2008 #9
    9.81= -w2 A cos(wt + teta)


    A=9.81/(-w2 A cos (wt + teta)

    i thought of this...

    what puzzle me...is how to calculate the

    cos (wt + teta) part....was the value of t and teta????
  11. Aug 20, 2008 #10


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    You had already determined that 9.81 was the maximum acceleration. Think about what range of values the cosine function is able to have; what then is the cosine equal to? In other words, when the acceleration is a maximum, what is the value of the cosine function?
  12. Aug 20, 2008 #11
    Omg!...on my way home from school...it struck me and i solve it...hurray hurray...

    9.81= -w2 A cos(wt + teta)


    A=9.81/(-w2 cos (wt + teta)

    at maximum amplitude the cos (0) which will give a value of 1.

    that will give us

    A=9.81/(-w2 (1) )

    Hurray! hurray!...finally got it...

    Thank you guys...
  13. Aug 20, 2008 #12


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    That's right. To take care of the minus sign, you might want to think of using [itex]\cos(\pi)=-1[/itex], which will give the same magnitude of the amplitude.
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