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Simple pendulum equations

  1. Dec 16, 2008 #1
    A simple pendulum has a period of 2 s for small amplitude oscillations.

    1. The length of the pendulum is most nearly
    (A) l/6 m
    (B) ¼ m
    (C) ½ m
    (D) 1 m
    (E) 2 m

    I used the period of a pendulum equation, did some algebra, and got D as the answer. Is this correct?

    2. Which of the following equations could represent the angle θ that the pendulum makes with the vertical as a function of time t?

    (A) θ = θmax sin (π/2)t

    (B) θ = θmax sin πt

    (C) θ = θmax sin 2πt

    (D) θ = θmax sin 4πt

    (E) θ = θmax sin 8πt

    I don't get this one. I'm assuming it's C because 2π = ω in a lot of equations and maybe it's true here too? Can someone please explain this?
  2. jcsd
  3. Dec 17, 2008 #2


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    1. Yes, good.

    2. No. One way to think about this is, after one period the argument of the sine should be 2π. And you know what t is after one period.
  4. Dec 17, 2008 #3
    I redid #2 this way.

    T = 2π/ω
    Substituted 2s for T and got ω = π

    So the answer would be B?
  5. Dec 17, 2008 #4


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  6. Apr 27, 2010 #5
    Could someone help me out for #2. I'm not even sure how to approach that part.
  7. May 9, 2010 #6
    bump, please help
  8. May 9, 2010 #7
    They solved it check the previous posts ..
  9. May 9, 2010 #8
    I saw the answer but I didn't understand why they are plugging in T. What exactly is the concept of θ = θmax sin Wt? I've never learned that so I don't know how to manipulate the equation.
  10. May 9, 2010 #9
    you can reach to the equation θ = θmax sin Wt by considering (i.e.) a pendulum of length l and start from newton`s 2nd law which states that net force = m x`` , with little work you will end up with the solution θ = θmax sin Wt, where W = sqrt(g/l) ..

    you can approach the same answer by substituting l = 1m, g=10m/s^2 in W , then you will get W = 3.16 for g = 10m/s^2 (or W=3.13 for g=9.8 m/s^2) which is approximately equal to the answer they reached .. :)
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