Pendulum in Accelerating Elevator 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"?
 A: 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 $g+a$.
A: If an object is accelerating upwards at a rate of $a$ m/s$^2$, then the gravitational force felt by this object is effectively,
$$
g_{eff}=g+a
$$
where $g\sim9.8$ m/s$^2$ is the canonical gravitational acceleration we all know and love.
In your particular case, the common equation of motion for a pendulum,
$$
\frac{d^2\theta}{dt^2}= - \frac{mg\sin\theta}{l}
$$
replaced $g$ with the effective $g$ and substituted:
$$
\frac{d^2\theta}{dt^2}= - \frac{mg_{eff}\sin\theta}{l}= -\frac{m(g+a)\sin\theta}{l}
$$
A: we can understand simply how the time period of a pendulum increase and decrease in an elevator.
1- when we go downward in an elevator(downwards accelerated elevator ) . we feel defect in our wight.
we know wight   $W=mg$.
$W$ our wight  is decreasing as elevator going downward . our mass is constant . so $g$ acceleration due to gravity is deceasing  .
time period of pendulum $T=2\pi\sqrt{\frac{l}{g}}$ 
as $g$ is decreasing in downward going elevator . 
SO time period of pendulum will increase.
2-opposite will take place  in upward accelerated elevator. time period pendulum will decrease.
3- if the wire of elevator break and elevator falling freely then $g=0$ time period will be infinite . pendulum will move in circular path.    
