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Simple Harmonic Motion problem

  1. Apr 24, 2012 #1
    1. The problem statement, all variables and given/known data
    35jgn74.jpg

    An elastic string is attached to A on a horizontal plane and a mass is attached to the other side of the rope.Then the rope is pulled to C and release.It engage in SHM from C to B and D to E and so on
    natural length = l
    extension = a
    max end points = E & C
    1. periodic time of the SHM ?
    2. Center if oscillation ?
    3. amplitude of the SHM ?


    2. Relevant equations


    3. The attempt at a solution
    I know that it doesn't engage SHM from B to D as the rope has not exceeded its natural length at that time...From B to D it travels at a constant velocity...As the periodic time of the oscillation is the time that it takes to complete an oscillation,I think it must be 2tEC,but not sure maybe its 2(tED+tBC).
    Actually I don't have an idea about the center if oscillation whether its A or B and D.
    And also amplitude.I think its DE or BC not AE as the SHM happens in only that area..
     
  2. jcsd
  3. Apr 24, 2012 #2

    gneill

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    Staff: Mentor

    Presumably the question would like you to consider the end-to-end motion of the mass (between C and E) to comprise the SHM. If that is the case, then what strategy would you propose for finding the period?
     
  4. Apr 24, 2012 #3
    I think that the period is 2tEC
    tEC=tAD+tDE
    to find tAD we can use SUVAT equations.
    and tDE=(∏/2)/ω
     
    Last edited: Apr 24, 2012
  5. Apr 24, 2012 #4

    gneill

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    Okay, but what then is ω? If the problem is cast in the form of spring-driven SHM (with an elastic string), perhaps you need to incorporate springlike properties in your solution? Define the mass M and the spring constant k and you should be able to derive the period in terms of k, M, l, and a.
     
  6. Apr 24, 2012 #5
    here ω is the angular velocity (dθ/dt)
    With F=ma and Hooke's law,I can get a function like f99ceb8a77f0ae8e0685bd0bc9985e32.png
    and calculate the period..any way the process is simple for me...But I want's to know,
    periodic time of the SHM
    Center if oscillation
    amplitude of the SHM

    Just please define them using letters in the image..

    optional:
    Code (Text):
    I'm doing this sum with a small use of differential calculus.Actually according to the Applied mathematics syllabus in our country...Any way..
    in the syllabus we don't have k,we use λ(modulus of elasticity)
    and hence k=λ/l
    thanks
     
    Last edited: Apr 24, 2012
  7. Apr 24, 2012 #6

    gneill

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    Staff: Mentor

    The period of anything with a cycle is the time to start from a given state and return to that state. In this case pick a point on the trajectory of the mass and (such as point C) and find an expression in terms of l, a, M, and k (or ##\lambda## if you prefer) for time for the mass to travel from C to E and back to C.

    For the center of oscillation I would need to know if there is a particular definition used in your coursework. The one I'm familiar with pertains to physical pendulums and is related to the Radius of Gyration for a pivoted pendulum. Otherwise I would make the obvious choice and say that it corresponds to the average position of mass during a period (point A).

    The amplitude of the oscillation should be obvious from the trajectory; what is the maximum displacement from the center of oscillation?
     
  8. Apr 24, 2012 #7
    Center of oscillation is defined as where acceleration is zero.So its B for C to B motion and D for E to D motion.
    And the amplitude is DE for DE motion and BC for BC motion..
    (as shown in examples of my coursework..)

    I don't know what it will be to the whole journey..And I think that we can't take such thing as whole journey but taking the two SHMs(BC & DM) separately maybe the right way... Am I correct..
     
  9. Apr 24, 2012 #8

    gneill

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    I really can't say for sure. That would seem to define the center of oscillation as the entire stretch from B to D.
     
  10. Apr 24, 2012 #9
    Can a center be a stretch ?
    anyway..I have some another question..
    We know at the center of oscillation the acceleration is zero (a=0)....
    no acceleration means no resultant force..If so...Is the center of oscillation ,a equilibrium position (a balanced position) ?
     
  11. Apr 24, 2012 #10

    gneill

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    I suppose so. You can place the mass anywhere between D and B with no motion and it won't move. Only when you stretch past D or B will there be potential energy infused into the system.
     
  12. Apr 24, 2012 #11
    What about the last question I asked(Is the center of oscillation ,a equilibrium position (a balanced position) ?)..Its not about the question we discussed above..
     
  13. Apr 24, 2012 #12

    gneill

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    I answered above :smile: If the mass doesn't move if it's placed there, the net forces on it must be zero -- hence it's in equilibrium.
     
  14. Apr 24, 2012 #13
    So can u answer this...?

    A mass is hanged by an elastic rope..So there's an extension in the rope and the mass is in its equilibrium(balanced) position(T=mg)...ok..
    now the mass is pulled down to some length from that balanced position..Then releases...
    Now it asks to find the time when it passes the equilibrium position...I want to know whether the T=mg position is a=0(the center of SHM) position or not ?
     
  15. Apr 24, 2012 #14

    gneill

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    I can answer it, but it is more important that YOU can answer it :wink: What's the net force acting on an object when it's "in equilibrium"? What's the acceleration of an object in equilibrium?
     
  16. Apr 24, 2012 #15
    got ya !
     
  17. Apr 25, 2012 #16
    (Sorry if I'm making trouble...)
    Is center of oscillation of any SHM ,in equilibrium ?
    We know that in the center of oscillation in SHMs of masses on a spring or a elastic string there's no any resultant force...so it in equilibrium..
    But when we consider about the SHM of a simple pendulum or a mass that engage in uniform circular motion....there's a resultant force at their center of oscillation(Centripetal force)....
    So pls make this clear...thanks !
     
  18. Apr 25, 2012 #17

    gneill

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    Staff: Mentor

    In these cases there is no force component that will change the speed of the mass at the equilibrium point. Centripetal forces do not act along the line of motion.
     
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