Spring-Block system executing SHM in a freely falling elevator

[vijaypandey93]

A block of mass m is suspended from the ceiling of a stationary elevator through a spring of spring constant k.Suddenly the cable breaks and the elevator starts falling.Show that the block executes S.H.M of amplitude mg/k in the elevator.

I solved the problem applying a pseudo force mg on the block in the opposite direction to that of the elevator. I would appreciate highly if someone can solve without applying pseudo forces.

My solution:-As in the equilibrium position,the spring is stretched by mg/k,when we apply mg opposite to that of the weight,it instantly cancels the weight and only force left is kx and that makes it execute in S.H.M and mg/k is the extension when the block is at rest w.r.t. the elevator and so the amplitude.Please solve without pseudo force.Many thanks in advance.

 

[tiny-tim]

welcome to pf!

hi vijaypandey93! welcome to pf!

… Please solve without pseudo force.

write out the equations of motion for the mass m, and for the top of the spring (where it is attached to the elevator) …

show us what you get ​

 

[vijaypandey93]

As the elevator is under free fall n spring is connected to it so i think,the top of the spring is also under free fall,and for the block:mg-kx=ma.a is the accleration of the block which is not equal to g because it's constrained by the spring force.what to do now?

 

[tiny-tim]

now write them out, using eg x₁ and x₂ instead of the same x for both!

 

[vijaypandey93]

bt then we can't have free fall of the ceiling because there'd be two forces working,one Mg(M being the mass of d ceiling) and the other kx1,both downward will get added to produce accleration more than g.what abt this?

 

[tiny-tim]

i suspect we're expected to ignore that!

(didn't you have to ignore it to use your pseudo force anyway?)

 

[vijaypandey93]

na,i don't think so because then i am in the elevator's frame of reference in which the ceiling is at rest.

 

[tiny-tim]

but if the mass m is not negligible, then the acceleration of your frame isn't constant

 

[vijaypandey93]

but in weightlessness spring doesn't apply ny force on the ceiling and can't we treat the free fall of all the bodies separately??

 

[tiny-tim]

but in weightlessness spring doesn't apply ny force on the ceiling

it jolly well does!

it's a spring

it's what springs do! ​

 

[vijaypandey93]

okay can you pls provide me the mathematical solution of it?

 

[tiny-tim]

okay can you pls provide me the mathematical solution of it?

erm …

noooo! ​

 

[mfb]

We do not provide full solutions here.

If you do not assume that the elevator has an "infinite" mass (large relative to m) the solution will be different. This is easy to see if you use a very small mass (<m, for example) and see how the system will look like.

 

[BruceW]

hehe. anyway, as tiny-tim was saying, it is very likely that you are supposed to ignore the sinusoidal motion of the elevator. So just assume the elevator is accelerating downward at g.

If you really want, you can take into account the SHM of the elevator, then take the limit that the mass of elevator is very large. You will get the same answer this way, but it will require more equations.

Do whichever way you are more comfortable with. In either case, you need to consider the motion of the ceiling and the motion of the block relative to the ceiling.

 

[vijaypandey93]

ok,can we be sure it executes SHM?i mean why does it execute SHM.when it's falling freely,the block is accelerating downward only because of the weight of it and the spring force is only applicable when it's stretched.so I assume that it's stretched initially,as the ceiling is also coming down with it,so can't that compression compensate the elongation??:P

 

[mfb]

Letting the elevator fall is like "switching gravity off" (that is the basis of General Relativity, by the way) - and the equilibrium position with gravity and without are different.

 

[vijaypandey93]

yeah i agree with you.bt how'd we show it mathematically?writing eqn for the block:mg-kx=mg will bring absurd results.what about this?can you guide me through solving my problem frm ground frame f reference.i'd really be thankful to you.

 

[WannabeNewton]

Note that your method of showing SHM in the elevator frame is the correct one and is the most elegant way to do it mathematically. You simply ascribe a pseudo force in the opposite direction to that of gravity and solve Newton's 2nd law in the elevator frame for the mass-spring attached to the ceiling of the elevator. If you are in the elevator frame you must take this pseudo force into account (as you have).

As for the ground frame, for simplicity assume the ground frame observer is directly above the elevator. For convenience choose the origin of the elevator frame as the relaxed length of the spring (which will be what?). Now write out Newton's 2nd law in the ground frame for the mass-spring. What position vectors will you have to consider?

 

[vijaypandey93]

For convenience choose the origin of the elevator frame as the relaxed length of the spring (which will be what?)

i'm sorry,i din't get wht u meant by chhosing the origin of elevator frame as the relaxed length of the spring.

 

[WannabeNewton]

Can you write out Newton's 2nd law in the ground frame for me? It will become clear after that.

 

[vijaypandey93]

Can you write out Newton's 2nd law in the ground frame for me? It will become clear after that.

okay,i try.for the block if we consider the acceleration a of d block downward, and if the elongation is x of d spring.then the eqn probably looks like:mg-kx=ma
one more thing we can do is to split this x into two parts as tiny tim said,x1 and x2 so x2-x1 =x if it works.x1 is d compression coz of the ceiling coming down and x2 the extension coz f the smaller block.n so the complete elongation is x=x2-x1.

 

[WannabeNewton]

Ok, I'm just having a bit of trouble understand your English but let's see what we can do. I think you have the right idea. Let's start with the basic form (I'm assuming you've taken the down direction to be positive based on your sign for ##mg##) ##m\ddot{x} = -k\Delta s + mg##. ##x## is the position of the block relative to the ground and ##\Delta s## is the displacement of the block from the equilibrium length of the spring (the elongation). If we let ##x_{\text{eq}}## represent the position of the equilibrium length of the spring relative to the ground then we can write ##\Delta s = x - x_{\text{eq}}## as you very well said yourself. Now how is ##x_{\text{eq}}## changing relative to the ground? Is it in one place or accelerating or what?

 

[vijaypandey93]

Ok, I'm just having a bit of trouble understand your English but let's see what we can do. I think you have the right idea. Let's start with the basic form (I'm assuming you've taken the down direction to be positive based on your sign for ##mg##) ##m\ddot{x} = -k\Delta s + mg##. ##x## is the position of the block relative to the ground and ##\Delta s## is the displacement of the block from the equilibrium length of the spring (the elongation). If we let ##x_{\text{eq}}## represent the position of the equilibrium length of the spring relative to the ground then we can write ##\Delta s = x - x_{\text{eq}}## as you very well said yourself. Now how is ##x_{\text{eq}}## changing relative to the ground? Is it in one place or accelerating or what?

the position of the equilibrium of the spring is not mentioned in the questions.the only thing mentioned is the elevator starts falling freely means tht ceiling of the elevator has an acceleration g in downward direction and we can't even say that the block will also freely fall.bcoz then it won't execute shm as evident from the equation mg-kx=mg.here m is the mass of the block and x is the elongation n if block is falling down with the same acceleration as its weight accounts for,then the restoring force because of spring will be vanished and there'd be no poimt of shm withouth the restoring force.so we simply can't have the acceleration of the block as g.

 

[WannabeNewton]

Just because it isn't mentioned doesn't mean you can't figure it out with the given information. I'm really having trouble understanding your text because you keep using text speak. Newton's 2nd law for the block in the ground frame reads, as mentioned above, ##m\ddot{x} = -k(x - x_{\text{eq}}) + mg## where ##x## is the position of the block relative to the ground and ##x_{\text{eq}}## is the position of the equilibrium length relative to the ground.

If the elevator was fixed in space then the equilibrium length would of course also be fixed in space. But for this problem, relative to the ground the elevator is in free fall so the position of the equilibrium length of the spring must be falling along with the elevator, do you agree? No one is saying that the block is in free fall here. So how is ##x_{\text{eq}}## changing relative to the ground?

 

[vijaypandey93]

Just because it isn't mentioned doesn't mean you can't figure it out with the given information. I'm really having trouble understanding your text because you keep using text speak. Newton's 2nd law for the block in the ground frame reads, as mentioned above, ##m\ddot{x} = -k(x - x_{\text{eq}}) + mg## where ##x## is the position of the block relative to the ground and ##x_{\text{eq}}## is the position of the equilibrium length relative to the ground.

If the elevator was fixed in space then the equilibrium length would of course also be fixed in space. But for this problem, relative to the ground the elevator is in free fall so the position of the equilibrium length of the spring must be falling along with the elevator, do you agree? No one is saying that the block is in free fall here. So how is ##x_{\text{eq}}## changing relative to the ground?

First of all I'm sorry.i''ll try not to use abbreviations anymore.anyways,yeah the equilibrium length is fixed because when the elevator was fixed in space,block was also at rest.means the downward acceleration was then 0.so mg-kx=0 and x=mg/k and x is actually x(equilibrium)(I don't know how're you typing the terms.so I'm writing like this only)

 

[WannabeNewton]

Ok good that is the magnitude of the equilibrium length itself but is the position of the equilibrium length fixed relative to the ground? The position of the equilibrium length is fixed relative to the elevator right? You agree with that? But the elevator is in free fall relative to the ground right? So how should the position of the equilibrium length relative to the ground also be changing?

As for typing the symbols, there's a FAQ on this forum about how to use Latex. Let me try to find the link and I will edit it into this post.

 

[vijaypandey93]

And one more thing,we only know the elevator is falling freely(given) and the spring is attached to the elevator.so at that point spring's acceleration too should be considered g only.do u agree?

 

[vijaypandey93]

Ok good that is the magnitude of the equilibrium length itself but is the position of the equilibrium length fixed relative to the ground? The position of the equilibrium length is fixed relative to the elevator right? You agree with that? But the elevator is in free fall relative to the ground right? So how should the position of the equilibrium length relative to the ground also be changing?

As for typing the symbols, there's a FAQ on this forum about how to use Latex. Let me try to find the link and I will edit it into this post.

hmmm..i agree.Equilibrium position is fixed relative to the elevator.And if elevator's acceleration is g relative to earth,then equilibrium position should also have an acceleration g relative to earth.right?

 

[WannabeNewton]

Perfect! You got it! Ok so let's go back to ##m\ddot{x} = -k(x - x_{\text{eq}}) + mg = -kx + mg + kx_{\text{eq}}##. Now you just concluded that ##\ddot{x}_{\text{eq}} = g## right? So ##x_{\text{eq}}## will be changing with time in the ground frame just like a regular particle dropped from rest at some height that subsequently falls due to gravity right? So in the ground frame, can ##x## describe just simple harmonic motion (i.e. will ##x## be a purely sinusoidal function)?

 

[vijaypandey93]

Perfect! You got it! Ok so let's go back to ##m\ddot{x} = -k(x - x_{\text{eq}}) + mg = -kx + mg + kx_{\text{eq}}##. Now you just concluded that ##\ddot{x}_{\text{eq}} = g## right? So ##x_{\text{eq}}## will be changing with time in the ground frame just like a regular particle dropped from rest at some height that subsequently falls due to gravity right? So in the ground frame, can ##x## describe just simple harmonic motion (i.e. will ##x## be a purely sinusoidal function)?

But do we've any relation between x(equilibrium) and x?i guess no.then how can we say that x'd be executing SHM in ground frame.i'm sorry but I din't get this point.

 

[WannabeNewton]

If ##\frac{\mathrm{d} ^{2}x_{\text{eq}}}{\mathrm{d} t^{2}} = g## then what is ##x_{\text{eq}}(t)## if the initial height of ##x_{\text{eq}}## above the ground was some value ##d## and it started off at rest before going into free fall?

 

[vijaypandey93]

If ##\frac{\mathrm{d} ^{2}x_{\text{eq}}}{\mathrm{d} t^{2}} = g## then what is ##x_{\text{eq}}(t)## if the initial height of ##x_{\text{eq}}## above the ground was some value ##d## and it started off at rest before going into free fall?

Then xequili(t)=g.(t^2)/2 + d

 

[WannabeNewton]

Then xequili(t)=g.(t^2)/2 + d

Right so plugging back into Newton's 2nd law we have ##m\ddot{x} = -kx + mg + k(\frac{1}{2}gt^{2} + d)## in the ground frame. Is this the equation for simple harmonic motion? If you solve this will you get an ##x(t)## that is purely sinusoidal?

 

[vijaypandey93]

Right so plugging back into Newton's 2nd law we have ##m\ddot{x} = -kx + mg + k(\frac{1}{2}gt^{2} + d)## in the ground frame. Is this the equation for simple harmonic motion? If you solve this will you get an ##x(t)## that is purely sinusoidal?

we can't say that necessarily because simple harmonic motion is caused by a force which has a restoring effect and opposite to the displacement.so all the terms may get added and may give a positive value which is ofcourse not the hall mark of SHM.

 

[vijaypandey93]

Right so plugging back into Newton's 2nd law we have ##m\ddot{x} = -kx + mg + k(\frac{1}{2}gt^{2} + d)## in the ground frame. Is this the equation for simple harmonic motion? If you solve this will you get an ##x(t)## that is purely sinusoidal?

okay i'll solve and then check it.thanks a lot