Accelerated Pendulum Frequency: g-a vs g+a

[KFC]

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

\ddot{\theta} + \frac{g}{l}\theta = 0

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

\omega = \sqrt{\frac{g-a}{l}}

why in textbook, in said the frequency becomes \omega = \sqrt{\frac{g+a}{l}}?

 

[sai_2008]

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?

 

[KFC]

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?

Yes, it makes sense.