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The question I have is similar to that of the Pendulum in an accelerating train problem. Where a bob is hung from the ceiling of a train that is at rest. The train then begins moving with an acceleration "a".

I understand that the mass of the pendulum bob does not affect the period of the swing. This fact can be verified using the formula to determine the period of a pendulum undergoing simple harmonic motion.

Will there be time delay between when the train first begins to accelerate and when the pendulum bob first begins to move, using a heavier bob compared to a lighter bob?

Hari
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  • I hope the question is clear. If not do let me know and I will attempt to phrase it better. – Hari Feb 23 '21 at 02:44
  • I don't understand. Are you asking if the period of the pendulum is affected if the train accelerates because the mass is decreased? Or do you mean if the time it takes to move (because of its inertia) changes if you decrease the mass? – joseph h Feb 23 '21 at 03:04
  • @joseph h I think the OP means the latter since it says "... bob first begins to move". – John Darby Feb 23 '21 at 03:15
  • Got it. I'll address a couple of scenarios since OP is not answering my question. – joseph h Feb 23 '21 at 03:33
  • @josephh I was referring to the time it takes to move due changes if you decrease the mass – Hari Feb 23 '21 at 04:58
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    @Hari OK. Well in that case there is no change in the time - they are both instantaneous. But I have written answers to both below. Cheers. – joseph h Feb 23 '21 at 05:03
  • of course, the pendulum's bob does not react to the train movement. It only reacts to its tether point being dragged to the side, inertia and gravity That's the whole point of being a pendulum. – PcMan Feb 23 '21 at 07:19

3 Answers3

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If you are asking how quickly the mass will respond to the train's acceleration, then the answer is instantaneously and this is independent of the mass of the bob. This is due to inertia and inertia is a property of all masses.

The period of a pendulum if it's in a stationary frame of reference is given by $$T = 2\pi \sqrt{\frac{l}{g}}$$ and the mass does not affect $T$ as you stated.

If you want to know if there is a difference in the value for $T$ depending on $m$ if the train accelerates, you must first ask if the motion of the train causes a dependence of $T$ on $m$:

Procedure:

If the trains acceleration is $a$, then the new acceleration of the bob, or its effective acceleration will be $$a' = \sqrt{g^2+a^2}$$ and you can get this by drawing vectors for $g$ and $a$ and use Pythagoras's theorem. This means that the new period will be $$T = 2\pi \sqrt{\frac{l}{a'}} = 2\pi \sqrt{\frac{l}{(a^2+g^2)^{\frac{1}{2}}}}$$

Again, even though there is an acceleration, the period of the pendulum is still independent of the mass of the bob. So the answer to your question must be no. There is no difference.

joseph h
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  • @ joseph h Suggest you comment on the answer by @Rishab Navaneet. Maybe you can explain it better than I did with my comment. I do not want the OP to be confused. – John Darby Feb 23 '21 at 04:21
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In the non-inertial reference frame that is the accelerating train, the bob experiences a fictitious force $-ma$ where $m$ is the mass of the bob and $a$ the acceleration of the train. This force is proportional to the mass of the bob. Assuming the bob is initially hanging vertically, as soon as the train accelerates, the initial displacement of the bob in the $-a$ direction is instantaneous and is the same regardless of the mass, since the force is constant and proportional to the mass. (This is similar to two objects of different mass dropped from a height; the objects travel the same distance in the same time- neglecting air resistance- since the force of gravity depends on the mass.)

For the pendulum motion in the accelerating frame, see the answer by @joseph h and also Time period of a simple pendulum in an accelerated frame on this exchange.

The following was added based on comments.

Assume the bob hangs vertically before the train starts to accelerate. When the train starts to accelerate there is instantaneous relative movement of the bob relative to the train seen by observers both on the train and on the ground. To the observer on the train, the train is not moving and the bob moves due to the fictitious force. To the observer on the ground, the bob is stationary and the train moves.

In both cases the relative motion of the bob relative to the train is the same and the bob continues to move relative to the train until a component of the tension force (a) counters the fictitious force for the observer on the train and, equivalently, (b) equals the acceleration of the train for the observer on the ground. The bob moves until it reaches a new equilibrium position. If the bob is "tapped" (slightly displaced from equilibrium), it acts as a pendulum about the equilibrium position. See Time period of a simple pendulum in an accelerated frame.

John Darby
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Previous answers are wrong in explaining when the bob starts to move... It is not instantaneous.

The bob is hung from the ceiling and it is the ceiling that is accelerating. The top portion of the string accelerates along with the train... But the portions of string down the line will start to displace only when that disturbance reaches them. More like when we create a wave on a string, the farther end displaces only when the wave reaches there.

Thus, the disturbance produced at the top end of the string takes some time to propagate down, characterised by the speed of transverse waves in that medium.

Now the speed of transverse wave in a medium is dependent on the tension in the string which in turn increases when you keep a heavier bob.

$$v=\sqrt{\frac{T}{\mu} } $$ where $\mu$ is the mass per unit length of string.

So if mass of the bob is increased, the tension in string increases, the waves travel faster and bob starts to move quicker

Rishab Navaneet
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  • *Assuming it is a real world scenario where your string is not massless. If it is a massless string, the displacement would be instantaneous. – Rishab Navaneet Feb 23 '21 at 03:50
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    The entire reference frame is accelerating and every object in that frame instantaneously experiences the ficticious force. See Goldstein, Classical Mechanics. The bob does start to move instantaneously regardless of the mass of the string, as does every other object in the accelerating frame, such as a person sitting in a chair. – John Darby Feb 23 '21 at 04:02
  • The entire reference frame does accelerate but the bob and string is not a part of that reference frame. The bob and string will remain inertial as long as the force reaches it. Except for the top most part of string that is attached to the train. We don't consider this delay because it is a very short time period and for low mass strings, negligible. Maybe that's why it is not considered. All calculations about the resting position of mass, its time period etc gives same ans if we consider or not consider this. – Rishab Navaneet Feb 23 '21 at 04:06
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    I think there's some misunderstanding here. Others (John Darby etc) are talking about when and how the bob starts to accelerate as measured in the train's frame. You (Rishab Navaneet) are talking about when and how the bob starts to accelerate as measured in an inertial frame. – g s Feb 23 '21 at 04:08
  • @g s. Yes that could be true... I have explained what happens for an outside observer. Someone inside would see the bob move back instantaneously and the string wiggle a bit – Rishab Navaneet Feb 23 '21 at 04:10
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    @RishabNavaneet It has nothing to do with the string and everything due to with the mass's inertia. You are also using an expression for velocity and we are talking about acceleration. – joseph h Feb 23 '21 at 04:27
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    @g s The bob moves instantaneously whether watched by an observer in the train (non-inertial frame) or an observer on the ground (inertial). – John Darby Feb 23 '21 at 04:35
  • See https://physics.stackexchange.com/questions/580273/explain-the-stationary-position-of-the-bob-from-ground-frame-of-reference on this exchange for the motion seen by an observer on the ground. – John Darby Feb 23 '21 at 04:57
  • @JohnDarby umm, no. if The bob does start to move instantaneously regardless of the mass of the string, as does every other object in the accelerating frame was true, then the pendulum would not swing – PcMan Feb 23 '21 at 07:22
  • @ PcMan The top of the string is connected to the train and is constrained to not move with respect to the train (the force of connection counters the fictitious force). The rest of the string and the bob experience the fictitious force and the bob moves sideways to a new equilibrium position. If the bob is "tapped" (slightly displaced from equilibrium), it acts as a pendulum about the equilibrium position/ See https://physics.stackexchange.com/questions/521229/time-period-of-a-simple-pendulum-in-an-accelerated-frame – John Darby Feb 23 '21 at 13:23
  • The relevant forces in our inertial frame are tension and gravity, initially equal and opposite, resulting in no acceleration. Gravity didn't change. Tension changed at the ceiling. We know how tension waves propagate along a wire. When the tension changes at the bob, the bob will accelerate. Everybody agrees that there is an instantaneous $d_2s/dt$ between any given point on the train and the pendulum bob, but from an inertial frame, it is the train that is accelerating while the bob is at constant velocity. The bob can't accelerate until the changed tension reaches it. – g s Feb 23 '21 at 18:33
  • that was supposed to be $d^2s/dt^2$ – g s Feb 23 '21 at 18:55
  • We are in agreement. My earlier comments were poorly worded. When train starts accelerate there is instantaneous relative movement of the bob and train seen by an observer (obs.) both on train and on ground. To the observer (obs.) on train, train is not moving and bob moves due to the fictitious force. To obs. on ground bob is stationary and train moves. In both cases relative motion of bob and train continues until the component of tension (a) counters the fictitious force for train obs. and (b) equals the acceleration of the train for ground obs. Sorry for the confusion. – John Darby Feb 23 '21 at 19:27
  • Also, the time for tension to propagate along the wire/rope/rod connecting the bob to the train is very short compared to the time for the bob to move any appreciable distance due to inertia. For this reason, physics mechanics texts treat problems involving tension- such a motion of a pendulum- assuming the tension force instantaneously changes as required when evaluating motion using forces/torques. – John Darby Feb 23 '21 at 20:57