Steel ball oscillating in a tube

  • #1
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Homework Statement:
a) Find the frequency of vibration under adiabatic conditions of a column of gas confined to a cylindrical tube, closed at one end, with a well-fitting but freely moving piston of mass m.

b) A steel ball of diameter 2 cm oscillates vertically in a precision-bore glass tube mounted on a 12-liter flask containing air at atomospheric pressure. Verify that the period of oscillation should be about 1 sec. (Assume adiabatic pressure change with γ = 1.4, Density of steel = 7600
Relevant Equations:
All below.
The problem is easy to solve, the question i have is another about static.
Why when we get:
F = (-A*p*γ/l)y
Can't we just substitute p*a = m*g? If this is a oscillation, it will be about some equilibrium position, where the net force is zero and which was the initial position of the body, in such position, p*a = m*g, but why this is wrong?
 

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  • #2
haruspex
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Homework Statement:: a) Find the frequency of vibration under adiabatic conditions of a column of gas confined to a cylindrical tube, closed at one end, with a well-fitting but freely moving piston of mass m.

b) A steel ball of diameter 2 cm oscillates vertically in a precision-bore glass tube mounted on a 12-liter flask containing air at atomospheric pressure. Verify that the period of oscillation should be about 1 sec. (Assume adiabatic pressure change with γ = 1.4, Density of steel = 7600
Relevant Equations:: All below.

The problem is easy to solve, the question i have is another about static.
Why when we get:
F = (-A*p*γ/l)y
Can't we just substitute p*a = m*g? If this is a oscillation, it will be about some equilibrium position, where the net force is zero and which was the initial position of the body, in such position, p*a = m*g, but why this is wrong?
Without seeing this in context of your whole solution, it's a bit hard to be sure I have understood your question.
In F = (-A*p*γ/l)y, if y is displacement from equilibrium position then F is net force, i.e. after taking mg into account. But if y is displacement from atmospheric pressure position then F is only the net force exerted by air.
 
  • #3
LCSphysicist
636
153
Without seeing this in context of your whole solution, it's a bit hard to be sure I have understood your question.
In F = (-A*p*γ/l)y, if y is displacement from equilibrium position then F is net force, i.e. after taking mg into account. But if y is displacement from atmospheric pressure position then F is only the net force exerted by air.
It is the first case, y is the net force, basically we came to this equation by:

PV^γ = Cte, so dp/P + γdv/V = 0 (1)

And the net force is F = dp*A, so we just just (1) here. But, as i said, p*A would need to be equal the weight, but for some reason it is wrong.

Essentially my doubt is that F = (-A*p*γ/l)y and F = (-m*g*γ/l)y would need to lead us to the same frequency and results, since P*A = M*g
 
  • #4
haruspex
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It is the first case, y is the net force, basically we came to this equation by:

PV^γ = Cte, so dp/P + γdv/V = 0 (1)

And the net force is F = dp*A, so we just just (1) here. But, as i said, p*A would need to be equal the weight, but for some reason it is wrong.

Essentially my doubt is that F = (-A*p*γ/l)y and F = (-m*g*γ/l)y would need to lead us to the same frequency and results, since P*A = M*g
Please clarify the meaning of some of these variables. In PVγ, P would be absolute pressure, but in P*A = M*g it would be gauge pressure at equilibrium, right?
dp is a small deviation in pressure, so can be relative to absolute or gauge, but what exactly is p?
 

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