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!

Accelerated pendulum

  1. Dec 26, 2008 #1

    KFC

    User Avatar

    Assume a pendulum, a mass M attached to a massless rod, which is oscillating in small angle. According to the oscillating equation

    [tex]\ddot{\theta} + \frac{g}{l}\theta = 0[/tex]

    We find that the frequency is proportional to [tex]\sqrt{g/l}[/tex]. Now if the pendulum is accelerated upward at acceleration [tex]a[/tex], since [tex]a[/tex] is along the opposite direction as the gravity [tex]g[/tex], the frequency of the accelerated pendulum should be

    [tex]\omega = \sqrt{\frac{g-a}{l}}[/tex]

    why in textbook, in said the frequency becomes [tex]\omega = \sqrt{\frac{g+a}{l}}[/tex]?
     
  2. jcsd
  3. Dec 26, 2008 #2
    Aha! Actually, it is not the pendulum that is accelerated upwards! It is the peg that is holding the pendulum that is accelerated. So the pendulum itself experiences a pseudo force downwards.

    If you stand in an elevator and it accelerates at "a" then your weight
    as you would experience will be m(g+a), because ma downwards is the pseudo force.

    Pendulum is no different: the effective downward acceleration of the pendulum of mass m is (mg + pseudo force)/m = (mg + ma)/m = (g+a).

    Makes sense?
     
  4. Dec 26, 2008 #3

    KFC

    User Avatar

    Yes, it makes sense.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook