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ANLSCOL2.HLP
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1990-05-02
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24 lines
┌────────────────────────────┐
│ T = 2 * pi * sqrt( L/g ) │
└────────────────────────────┘
Reducing the oscillation amplitude with F3 doesn't change T either.
While the maximum and average velocities are reduced, the length of the
trajectory is also reduced and (for small amplitudes) in the same
proportion.
Notice that when you double L or halve g (by using F4 or F5) the
acceleration is halved. As you can see, in the former case this is due to
the angle changing and in the latter to the weight changing. But T (and
the average velocity) only change by the square root of 2. The reason is
the following: While the rate of change of the velocity is reduced to half
its former value, at each point of the trajectory, the velocity has now
more time to build up. This fact partly compensates for the reduced
acceleration. This is a qualitative explanation for the square root (sqrt)
in the above formula.
Use F6 to stop the action and to read the instantaneous values of
relevant quantities. In particular notice the values of the kinetic and of
the potential energies and of their sum, at different points of the
trajectory. What happens to these values when the pendulum is close to an
extreme position or when it goes through the equilibrium position?