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Harmonic motion of a box of mass

  1. Jun 9, 2012 #1
    (a) A box of mass m, attached to a spring, oscillates in simple harmonic motion on a frictionless surface as indicated in the figure below. Each time the box reaches the point of maximum displacement a child allows some sand to fall vertically into the box. This will gradually increase the mass of the system. Briefly explain what will happen to the period of oscillation as sand is added.

    (b) A spring hangs vertically from a ceiling with a mass m = 3.0
    kg attached to its lower end. The spring constant is k = 45 N.m-1.
    Initially, you hold the mass in your hand so that the spring is at its natural length (neither stretched nor compressed). At t = 0 s you release the mass from rest.

    (i) Find a value for the period of oscillation of this spring-mass system.

    (ii) Find a value for the amplitude.
     
  2. jcsd
  3. Jun 9, 2012 #2
    Hi Sydboydell31,
    You have to find the values and we will try to help you if you encounter any problems.
     
  4. Jun 9, 2012 #3
    so what's i've got so far is:
    a) i've proved that the period increases when mass is increased because of F=ma, the heavier the mass the bigger the acceleration, when the mass increases, more force is then able to be compressed into the spring and therefore transferred back. Since there's no friction in this question, there will be no loss of energy during the oscillation, so if the speed increases, the period decreases. is this right?

    b) (i)
    m=3.0kg
    k=45N.m-1
    t=0
    i've used the equation ω=sqr(k/m)=sqr(45/3)=3.87kgN-1m-1
    Then used the equation ω=2pi/T, to find the period, 3.87=2pi/T, T=1.62s

    c) (ii)
    i'm not too sure how to approach this question, do you use the equation ωA=(2pi)/T?
     
  5. Jun 10, 2012 #4
    my take:

    for a) the period wont change the period of pendulum = 2pi/w = 2pi(L/G)^0.5 so mass doesnt affect period of pendulum

    b) yes use the formula

    c) not sure
     
  6. Jun 10, 2012 #5
    Yep. You can also observe this from the formula,

    [tex]T = 2\pi \sqrt {\frac{m}{k}}[/tex]


    But this isn't a pendulum :wink:

    That seems correct to me.

    The formula you wrote requires does not give you amplitude... It should be,

    [tex]\omega = 2\pi/T[/tex]

    Amplitude is the maximum displacement from the mean position. How can you find the maximum displacement??
     
  7. Jun 13, 2012 #6
    Just read the forum FAQs. My bad.
     
    Last edited: Jun 13, 2012
  8. Jun 13, 2012 #7
    Just read the forum FAQs. My bad.
     
    Last edited: Jun 13, 2012
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