![]() ![]() So we have approximate simple harmonic motion, where w 2 = g/l. If the particle is at P or Q when t = 0, then the following equation also holds:Ī simple pendulum consists of a particle P of mass m, suspended from a fixed point by a light inextensible string of length a, as shown here: If the particle is at 0 when t = 0, then the following equation also holds: The periodof the motion is the time it takes for the particle to perform one complete cycle. Hence the maximum velocity is a w (put x = 0 in the above equation and take the square root). Where v is the velocity of the particle, a is the amplitude and x is the distance from O.įrom this equation, we can see that the velocity is maximised when x = 0, since v 2 = w 2a 2 - w 2x 2 We can solve this differential equation to deduce that: If so, you simply must show that the particle satisfies the above equation. You may be asked to prove that a particle moves with simple harmonic motion. Where w is a constant (note that this just says that the acceleration of the particle is proportional to the distance from O). The amplitudeof the motion is the distance from O to either P or Q (the distances are the same).Ī particle which moves under simple harmonic motion will have the equation The particle will therefore move between two fixed points (P and Q). It will keep going and then again slow down as it reaches P before stopping at P and returning to O once more.
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