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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"?

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  • $\begingroup$ What's do you mean by proving it? It's the thing of common sense.. $\endgroup$ – Schrödinger's Cat Nov 23 '14 at 3:37
  • $\begingroup$ Why did you try looking up a FBD? Have you considered drawing one yourself? Once doing so, it's straight-forward to see the $m(g+a)\sin\theta$ relation. $\endgroup$ – Kyle Kanos Nov 23 '14 at 3:40
  • $\begingroup$ I do not see why it is common sense... Why would the axial component of the acceleration have to be summed with masin($\theta$)? $\endgroup$ – yolo123 Nov 23 '14 at 3:40
  • $\begingroup$ @KyleKanos That is what I did. $\endgroup$ – yolo123 Nov 23 '14 at 3:41
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    $\begingroup$ Look up fictitious forces. If you do that, it's a one-step problem to understand how $mg\rightarrow mg+ma$. $\endgroup$ – BMS Nov 23 '14 at 4:17
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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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  • $\begingroup$ I lost you there at the last step: So, I do: Tcos(theta)-mg=m(y+a)? $\endgroup$ – yolo123 Nov 23 '14 at 4:02
  • $\begingroup$ Ah yes! I understand. $\endgroup$ – yolo123 Nov 23 '14 at 4:02
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    $\begingroup$ @yolo123 It's a relief when one gets over little hurdles like this eh (to me it feels like pulling a thorn out)? Don't pay any heed to the comments trivialising your problem, we all get stuck on silly things like this all the time (or at least I do). You're clearly new to Newtonian mechanics and you're asking the right questions to get yourself a sound understanding: well done. $\endgroup$ – Selene Routley Nov 23 '14 at 4:07
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    $\begingroup$ Thanks, this analysis was exactly what I was looking for. $\endgroup$ – yolo123 Nov 23 '14 at 4:29
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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: $$ \frac{d^2\theta}{dt^2}= - \frac{mg_{eff}\sin\theta}{l}= -\frac{m(g+a)\sin\theta}{l} $$

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  • $\begingroup$ don't you mean g = -9.8 m/s^2 ? $\endgroup$ – jsky Dec 22 '16 at 1:26
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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.

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