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Spring Problem

  1. Dec 12, 2016 #1
    1. The problem statement, all variables and given/known data
    You are in a space ship far from any other objects, and you want to build a clock. You decide to build your clock out of a spring with a mass attached to it. You use a spring with spring constant k = 138 N/m, and you initially displace the mass a distance x=25.0 cm from equilibrium.

    a) How much mass will you need so that your clock will measure seconds?
    b) What is the max velocity of the attached mass?
    c) Draw a plot of the spring potential energy (Usp) as a function of time for two periods of the clock. The plot must be neat and include numeric values for the max Usp. Make sure you also label the points when Usp is a min and a max, Give the numeric values for both the time and Usp.

    2. Relevant equations
    T= 2π ⋅ √m/k

    V(max) = ± A ⋅ √k/m

    3. The attempt at a solution

    a) T2/ 4π2= m/k
    m=(k * T^2)/(4 * π2)
    m= 138 * 1 sec / ( 4 * π^2)
    m= 3.50 kg (is this correct?)

    b) V (max) = ± 0.25 m * √(138 N/m /3.496 kg) = ± 1.57 m/s (is this correct?)

    c) I don't understand this part.
     
    Last edited: Dec 12, 2016
  2. jcsd
  3. Dec 12, 2016 #2
    It should be periodic, since you know the solutions for harmonic oscillators (x=A sin(wt+e)) and the spring potential energy (U(x)=1/2kx^2) and just substitute ,
    U(t)=1/2 k A^2 sin^2(wt+e) , now you can plot this function with numerical values.
     
  4. Dec 12, 2016 #3
    Can you write any equation for the potential energy? Seeing as you have determined the period and the amplitude can you write an equation for the displacement (x) as a function of time?
     
  5. Dec 12, 2016 #4
    you can find e using the initial condition
     
  6. Dec 12, 2016 #5
    what is e? The equation in my book just says x= A ⋅ cos (ω ⋅ t) and this goes to x = A ⋅ cos ( 2π/T ⋅ t)
    And how do you find t?
     
    Last edited: Dec 12, 2016
  7. Dec 12, 2016 #6
    In the case of his equation ##U(t)=\frac 1 2 k A^2 sin^2(\omega t+e)## e is the phase for the harmonic oscillator. It determines the starting point for the sin wave. You can find it using the equation ##x=A sin(\omega t+e)## since you know what x is at time zero.
     
  8. Dec 12, 2016 #7
    Ah I see. So how do you isolate e in this equation: ##x=A sin(\omega t+e)##? I don't remember how to do that where the variable is locked up in the sine function.
     
  9. Dec 12, 2016 #8
    Plug in your two known values for position and time. This will get you ##25cm=Asin(e)## since ##\omega t## goes to zero. Knowing how a sin wave behaves, and that 25 centimeters is your max amplitude, you should then be able to solve for e.
     
  10. Dec 12, 2016 #9
    well I did (0.25 m / 0.25 m) = sin(e) ----> 1=sin(e)
     

    Attached Files:

  11. Dec 12, 2016 #10
    Right, now take the inverse sin of both sides and you'll have e.
     
  12. Dec 12, 2016 #11
    ok thanks. I forgot about the inverse function.
    So e = 1.571
     
  13. Dec 12, 2016 #12
    So is this the correct plot then?

    ω= 2π / T ---> 1.000631908

    u(t) = 0.5 * 138 * (0.25)2 * sin^2(1.000631908 ⋅ t +1.571)
    simplifies to:
    u(t) = 4.3125 * sin^2 (1.000631908 ⋅ t +1.571)
     

    Attached Files:

  14. Dec 12, 2016 #13
    It looks right, except I'm confused about your value for ##\omega##. It should be equal to ##\frac {2\pi} T## where T is 1 such that it completes a full cycle every second. For the record, it's also usually easier (and more accurate) to keep things out of decimal form in your equations.
     
  15. Dec 12, 2016 #14
    Ah I see. I had an error with my T.
    ω should equal 2.51.

    Here's the new graph:
     

    Attached Files:

  16. Dec 12, 2016 #15
    Still not quite right. Your ##\omega## should equal ##2\pi## if you want it to make one full cycle (from 25cm to -25cm back to 25cm) every second. This isn't a calculated value in this case. The problem statement specifies what they want the period to be.
     
  17. Dec 12, 2016 #16
    Okay so the period is 1 rev/sec since it's a clock right?
    So ω= 2π/1 = 6.283185307

    Here is the graph:
     

    Attached Files:

  18. Dec 12, 2016 #17
    Precisely.
     
  19. Dec 12, 2016 #18
    I'm still confused, how did you get U(t)=1/2 k A^2 sin^2(wt+e). How did that substitution work?
     
  20. Dec 12, 2016 #19
    The potential energy of a spring is given by the equation ##U(t)=\frac 1 2 kx^2##. He subbed in the equation of x for a harmonic oscillator, ##x(t)=Asin(\omega t +\phi)##, into it.
     
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