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Oscillation (SHM)

  • #1

Homework Statement


Some sand is sprinkled onto the cone. The sand oscillates vertically with the frequency of the cone. The amplitude of oscillation of the cone is increased.
At a particular amplitude of oscillation the sand begins to lose contact with the cone.
By considering the forces acting on a grain of sand, explain why this happens.

Homework Equations


From mark scheme it says "Weight - contact force = m(w^2)x So as x increases, contact force decreases, sand loses contact with cone when contact force = 0"

But I don't understand why it is m(w^2)x, not -m(w^2)x? as I thought they take upward as negative and losing contact only happens at maximum upward displacement position.

The Attempt at a Solution

 
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Answers and Replies

  • #2
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From mark scheme it says "Weight - contact force = m(w^2)x So as x increases, contact force decreases, sand loses contact with cone when contact force = 0"

But I don't understand why it is m(w^2)x, not -m(w^2)x? as I thought they take upward as negative and losing contact only happens at maximum upward displacement position.
pl. make out a diagram of oscillations and then mark out the forces acting on sand particle ;perhaps then you can gigure out the contact forces and required centrepetal/centrifugal force .
 
  • #3
pl. make out a diagram of oscillations and then mark out the forces acting on sand particle ;perhaps then you can gigure out the contact forces and required centrepetal/centrifugal force .
Yes I did!
 
  • #4
963
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Yes I did!
then show it as attachment so that we can get to your problem
 
  • #5
then show it as attachment so that we can get to your problem
_k8xiwHxV0bG9Qm2Ts8oD_m3dnq_2E2pyOJ2Xu78HfRZMMWe_zcAIIcCDmaJA?width=660&height=326&cropmode=none.jpg
 
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  • #6
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@mystreet123
well the attachment fails to open -perhaps a template will be better so that it can easily open.
 
  • #7
@mystreet123
well the attachment fails to open -perhaps a template will be better so that it can easily open.
Can you open it now?
 
  • #8
963
213
  • #9
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at some amplitude of oscillations of the cone the sand particles leaves the contact with surface

At a particular amplitude of oscillation the sand begins to lose contact with the cone.
From mark scheme it says "Weight - contact force = m(w^2)x S
suppose N is the contact force so N= mg - mw^2.x at any displacement x so when contact force is zero the wt will be balanced by the term on the right hand ;
as i feel mg is down ward so mw^2 .x shoukd be upward.

or think in terms of F(harmonic osc.) = F(Weight) -N
what is wrong in the above picture of forces?
 
  • #10
at some amplitude of oscillations of the cone the sand particles leaves the contact with surface





suppose N is the contact force so N= mg - mw^2.x at any displacement x so when contact force is zero the wt will be balanced by the term on the right hand ;
as i feel mg is down ward so mw^2 .x shoukd be upward.
what is wrong in the above picture of forces?
But why I found N-mg = mw^2.x at highest position? if we take upward positive

Because when I use this formula, as x increases, N also increases, instead of becoming zero
 
  • #11
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mw^2.x at highest position?
in a harmonic oscillation the force which is always a restoring force acts opposite to the displacement

F(osc)= -constantx displacement
if displacement is +ve - force will be negative and just vice-versa
if displacement from mean position is negative then force will be in opposite direction.
will this clarify your picture?
 
  • #12
in a harmonic oscillation the force which is always a restoring force acts opposite to the displacement

F(osc)= -constantx displacement
if displacement is +ve - force will be negative and just vice-versa
if displacement from mean position is negative then force will be in opposite direction.
will this clarify your picture?
Yes understood thanks!
 
  • #13
963
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Yes understood thanks!
in physics forum if you give thanks then just like the post
we do not write explicitly
 

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