I have been looking for this for quite some time now. A simple pendulum behaves in SHM. Let's put that pendulum in an upward accelerating elevator. The component of the force that acts in SHM $(\text{mg}\sin\theta)$ still stays the same in my head.
However, websites and books tell me to use $m(g+a)\sin\theta$ where $a$ is the acceleration of the elevator.
I tried to look up Free Body Diagrams, but I can't find any for the case of accelerating frames. Can someone explicitly prove this without using the flimsy argument of "Think it's a noninertial frame with a new effective g"?
Answer
Well it depends on the context of your question. If you're being introduced to General Relativity, then you're just going to assume, in the spirit of the equivalence principle, that gravity and the acceleration cannot be told apart from the pendulum's standpoint, so the acceleration is obviously $a+g$.
If you need to do it from first principles in a Newtonian setting, draw a free body diagram of the bob. First, let's do the unaccelerated pendulum. On the FBD, if you resolve the tension in the thread holding up the bob $(-T\,\sin\theta,\,T\,\cos\theta)$ together with the weight $(0,\,-m\,g)$ into horizontal and vertical components, you get:
$$-T\,\sin\theta = m\,\ddot{x}$$ $$T\,\cos\theta - m\,g = m\,\ddot{y}$$
but now, if you do it again with the bob and thread system accelerating upwards with constant acceleration $a$, then the $y$-component of the acceleration measured relative to the "inertial" (in Newtonian gravity) frame stationary wrt the ground is $\ddot{y}+a$ whilst $\ddot{x}$ is unaffected. So now, put these back into the equations above, and you find you get the same as the first set but with $g$ replaced by $g+a$.
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