1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Archived Pendulum Motion (Simple Harmonic Motion)

  1. Nov 5, 2009 #1
    1. The problem statement, all variables and given/known data
    A 180 g mass on a 2.5m long string is pulled 6.6 degrees to one side and released. How long does it take for the pendulum to reach 2.7 degrees on the opposite side?


    2. Relevant equations

    omega = sqrt(g/L)
    theta = S/L
    theta(t) = A * cos(omega * t + phi)

    3. The attempt at a solution

    I am looking at this problem in two parts, one is the time it takes the pendulum to swing 6.6 degrees (0.12 rad) and the other is how long it takes for the pendulum to go the remaining 2.7 degrees (0.047 rad). I have:

    omega = sqrt(g/L) = 1.98 rad/s

    For time it takes to swing 0.12 rad:

    theta = S/L => s = L * theta => s = 0.3m = A

    theta(t) = A * cos(omega * t + phi) => 0.12rad = 0.3 * cos(1.98t) => t = 0.59s

    For time it takes to swing 0.047 rad:

    theta = S/L => s = L * theta => s = 0.118m = A

    theta(t) = A * cos(omega * t + phi) => 0.047rad = 0.118 * cos(1.98t) => t = 0.59s

    Now, I know for a fact the second part is wrong as the pendulum doesn't go nearly as far. For the first part I left the phase constant as 0 since the pendulum is being held to the far right of its motion, however for the second part I'm not sure what to do with the phase constant. I know that the phase constant should probably be negative as it is pendulum motion to the left, however I'm not sure how to find it. I think if I can find the phase constant for the second part, use it to find the time, and add the times together I should get total time.

    Any help is appreciated.
     
  2. jcsd
  3. Mar 4, 2017 #2
    It would be best to work with 3 significant figures everywhere (0.12 rad isn't particularly accurate).

    You write:
    theta(t) = A * cos(omega * t + phi)
    But theta is an angle and A is a distance in metres. . . .
    It's the angular displacement that behaves simple-harmonically (i.e. varies sinusoidally)--there's usually no need to convert to a linear distance. In fact you could leave those angles in degrees; mathematically, degrees make as much sense as metres in this context, and they save work.

    (Perhaps a simpler way to think about this part: What is the period of the motion? What fraction of that period is the time taken for the mass to swing from the extreme to the centre? --Think about what a sine or cosine curve looks like and what sorts of symmetry it has.)

    For the second part, I think you're making it a bit more complicated than it needs to be. Remember the mass is in continuous sinusoidal motion from the time it's released.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Pendulum Motion (Simple Harmonic Motion)
Loading...