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Homework Help: Simple harmonic motion in a rocket?

  1. Apr 20, 2004 #1
    A simple pendulum suspended in a rocket ship has a period T0. Assume that the rocket ship is near the earth in a uniform gravitational field. (If A and E are true, and the others are not, enter TFFFT).

    A) If the ship accelerates upward, the period decreases.
    B) If the ship goes into freefall, accelerating downward at 9.81 m/s2, the pendulum will no longer oscillate.
    C) If the ship moves upward with a constant velocity, the period decreases.
    D) If the mass of the pendulum doubles, the period increases.
    E) If the length of the pendulum is doubled, the new period will be the square root of two times T0.

    I thought I had the answer on this one. But I didn't..

    My instinct was

    I also tried, TTTFT, TFTFT and FTTFT (ABCDE)

    As you can see I was very confident that C=T D=F E=T.. And I was wrong.

    Anyone have a better idea?
  2. jcsd
  3. Apr 20, 2004 #2
    So what i think now is XXFFT

    Really not sure on the first two, but i think that C=F D=F E=T
  4. Apr 20, 2004 #3
    I'm most probably wrong but I think the answer is given to you at the beginning there.
    So far I've come up with TFFFT.
    When the ship accelerates upwards won't that increase the value of the downward acceleration due to gravity? So according to the equation
    T=sqrt(l/g) (basic equation) wouldn't it cause the period to decrease.
    And with the last one I'm think i'm sure (haha) that it is right because if you do the maths it works out as square root of 2 times T0.
    I'm pretty sure that the second one is false.
    The third one is false as well I think, because if velocity is constant then acceleration is 0. So nothing will happen to the period.
    As for D, increasing the mass would decrease the period most likely.
    Hope that helps ya ;)
  5. Apr 20, 2004 #4
    K...you guys....remember this equation for the period of a pendulum when solving....

    T = 2(pi) * (length/g) ^ (1/2)....that means that when.....

    g increases.......the period decreases........
    ...accelerating upwards means that g will be increased on the ship.....moving upwards at a constant velocity has NO EFFECT whatsoever on g....I believe that the correct answers are....
    A. T (Yes....accelerating upwards will increase g...thus decreases T)
    B. T (yes....SHM requires a restoring forces...if the pedulum is in free fall...there is NO restoring force...)
    C. F (no...when it is moving at a constant velocity....the value of g is not affected)
    D. F (no..the mass of a pendulum as no effect on its period...as you can see from the equation)
    E. T (yes....the equation proves this to be the case)
  6. Apr 20, 2004 #5
    Cool thanks for the info on the restoring force man ;)
  7. Apr 20, 2004 #6
    Thanks alot for the explanations, you were right on :)
  8. Apr 27, 2004 #7
    np man...anytime :-)
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