Spring-Block system executing SHM in a freely falling elevator

  • #26
WannabeNewton
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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.
 
  • #27
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?
 
  • #28
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?
 
  • #29
WannabeNewton
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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)?
 
  • #30
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.
 
  • #31
WannabeNewton
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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?
 
  • #32
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
 
  • #33
WannabeNewton
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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?
 
  • #34
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.
 
  • #35
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
 
  • #36
WannabeNewton
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You don't have to explicitly solve it if you don't want to; you can tell just by looking at the equation.
 

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