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Vertical Simple Harmonic Motion

  1. Feb 26, 2015 #1
    1. The problem statement, all variables and given/known data

    ?temp_hash=e7d7ec91ba1db6928646ef6a7a3e9b65.png

    2. Relevant equations


    3. The attempt at a solution

    Suppose mass of platform is ##M_1## and that of coin is ##M_2## . The normal force between them is N .

    EOM for platform (mass M1) = ##-kx+M_{1}g+N = M_{1}\ddot{x}##

    EOM for coin(mass M2) = ##M_{2}g-N = M_{1}\ddot{x}##

    This gives , ## N = \frac{kxM_2}{M_1+M_2} ## . The normal force vanishes when x= 0 ,i.e when the spring acquires its natural length .

    I feel there is something wrong with my work . I would be grateful if somebody could help me with the problem.
     

    Attached Files:

  2. jcsd
  3. Feb 26, 2015 #2

    Suraj M

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    More than one option right?
    You start,!! you think Option A is right? or B?
    Also i think there might be a simpler way to solve you're problem apart from your posted method, just use ##a = \omega^2 x## and a=g if you get the first part (option A or B ) you'll know what is x.
     
  4. Feb 26, 2015 #3
    Hi Tanya. You really don't have to include the mass of the platform in the analysis. It is forcing the coin to move in the manner it tells the coin to move. So, the motion of the coin is just

    y = Asinωt

    From this, what is the acceleration of the coin? What does the force balance on the coin give you? Look at the force balance at 4 times through the period.

    Chet
     
  5. Feb 26, 2015 #4
    Hi Chet :oldsmile:

    Thanks for replying . I think I misunderstood the setup .

    Do you agree that both (1) and (3) are correct options ?
     
  6. Feb 26, 2015 #5
    Let us change the setup a bit .

    Instead of a platform undergoing SHM at constant angular frequency ##\omega ## , we have a block M1 attached to a vertical spring and block M2 placed on it . Both the blocks are pushed down a small distance 'd' and released . Now if we have to find the position and time where the upper block M2 leaves M1 ,then do you think my analysis is correct in the OP?

    Does the upper block leave the lower block when the spring returns to its unstretched length ?
     
  7. Feb 26, 2015 #6
    Yes.
     
  8. Feb 26, 2015 #7
    Okay .

    What is your opinion on post#5 ?
     
  9. Feb 26, 2015 #8
    I think it's not correct. I get the same result as with the original problem statement, except with ω^2=k/(M1+M2). For one thing, your force balance on M2 is incorrect.

    Chet
     
  10. Feb 26, 2015 #9
    What is the error ?

    Yes , but till the point they move together . Isn't it ? How would we find the position when M2 leaves M1?

    It is typo in EOM of M2 . It should be M2 on right side .
     
  11. Feb 26, 2015 #10
    I don't know where you get your algebraic equation for N.

    If I add your two equations together, I get:

    $$-kx+(M_1+M_2)g=(M_1+M_2)\frac{d^2x}{dt^2}$$
    From this we see that the equilibrium position of the spring is at:

    $$x_0=\frac{(M_1+M_2)g}{k}$$

    If we define y as the distance relative to the equilibrium position, then ##x=y+x_0##. If we substitute these into the differential equation, we get:

    $$-ky=(M_1+M_2)\frac{d^2y}{dt^2}$$

    The solution to this equation subject to your prescribed initial conditions is:
    $$y=Acosωt$$
    where ##ω^2=\frac{k}{(M_1+M_2)}##

    The acceleration is then ##y''=-Aω^2cosωt##
    If we substitute this into the equation for M2, we obtain:
    $$N=M_2(g+Aω^2cosωt)$$
    The right hand side of this equation has a minimum value when cosωt=-1. That's when the two masses are at their highest point.

    Chet
     
  12. Feb 26, 2015 #11
    Thanks Chet for the effort you have put in to help me understand the problem .

    But , instead of eliminating N as in the way you have done , if I eliminate ##\ddot{x}## from the two EOM's ,don't we get the normal N as a function of x i.e N(x) = ##
    N = \frac{kxM_2}{M_1+M_2}
    ## ?

    From this we get x = 0 , when N=0 .

    Please give your views .
     
  13. Feb 26, 2015 #12

    SammyS

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    I don't understand your sign conventions.

    Ignoring that, your "EOM" for mass, M2 in apparently incorrect.

    The right hand side should be, ##\ M_2\ddot{x}\ .## That is the result of the net force on the coin.
     
  14. Feb 26, 2015 #13
    x = 0 is not the equilibrium position. x = 0 is the unstretched length of the spring. It turns out that this is also equal to the highest position of the masses corresponding to the lowest value of A at which separation occurs.

    Chet
     
  15. Feb 26, 2015 #14
    When did I say that x= 0 is the equilibrium position :smile: ? But then what I found in OP that the blocks separate when the spring regains its natural length is correct .

    Right ?

    How did you get that ?

    From the equiibrium position , we push the blocks by a distance 'd' downwards ? Isn't 'd' the amplitude of the SHM ?
     
  16. Feb 26, 2015 #15
    From my previous analysis,

    x = x0 + y

    with ##x_0=\frac{(M_1+M_2)g}{k}##

    So,
    $$x=\frac{(M_1+M_2)g}{k}+y$$

    I also showed that, at the minimum amplitude for separation,

    $$y=-A=-\frac{g}{ω^2}=-\frac{(M_1+M_2)g}{k}$$

    So,
    $$x=\frac{(M_1+M_2)g}{k}-\frac{(M_1+M_2)g}{k}=0$$

    So basically, what this says is that you have to pull the masses down to at least 2X the equilibrium displacement in order for separation to occur. And you showed that if you pull the masses down at least that far, the separation will always take place when the spring attains its unextended length.

    Chet
     
  17. Feb 27, 2015 #16

    ehild

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    The platform is made to perform SHM with angular frequency ω and with increasing amplitude. That means its position is Asin(ωt), a known function of time, with A slowly increasing with time.

    You meant ##M_{2}g-N = M_{2}\ddot{x}## did you not?
    That is correct if you choose "down" positive. But the platform can exert only upward normal force, so it must be positive. The coin stays on the platform and does the same motion as the platform. x for the coim is the same as x for the platform. The acceleration of the platform is ##\ddot x = -\omega^2 x##, so
    ##N =M_{2}g- M_{2}\ddot{x}=M_{2}(g+\omega^2 x)## . At the top, x = -A, negative. The coin loses contact with the platform if the normal force needed to move it together with the platform should be negative. What does it mean for the amplitude?
     
  18. Feb 27, 2015 #17
    Sorry I don't understand this statement . What I have understood is that , only if we push the two blocks a minimum distance ##\frac{g}{ω^2}## , then only the upper block loses contact and that too at the upper extreme location ( i.e when the spring attains its natural length ) .
     
  19. Feb 27, 2015 #18
    Hello ehild :oldsmile:

    It means ## A ≥ \frac{g}{ω^2}## . Right ?
     
  20. Feb 27, 2015 #19

    ehild

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    Yes, And at what position?
     
  21. Feb 27, 2015 #20
    What I'm saying is that, if you start the masses out out at 1X their equilibrium displacement (that is at the equilibrium displacement below the unstretched spring position), you need to lower the masses to 2X the equilibrium displacement (i.e., an additional 1X the equilibrium displacement below the equilibrium displacement) before the mass M2 can begin to pop up after you release the masses. If you lower it to more than this location before you release, the coin will always pop up, and the location that it will do so is always when the spring attains its natural length (as you showed), which is at or below where it attains its upper extreme location.

    I hope this makes sense.

    Chet
     
    Last edited: Feb 27, 2015
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