Steel ball oscillating in a tube

In summary: Greek letter gamma, which is dimensionless. It is related to the speed of sound in a material and is used in equations to describe the effects of pressure on oscillations.
  • #1
LCSphysicist
646
161
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?
 
Physics news on Phys.org
  • #2
LCSphysicist said:
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
haruspex said:
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
LCSphysicist said:
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?
 

Related to Steel ball oscillating in a tube

1. What is the purpose of a steel ball oscillating in a tube?

The purpose of a steel ball oscillating in a tube is to demonstrate the principles of simple harmonic motion and energy conservation. It is also used to study the effects of different variables, such as mass and amplitude, on the oscillation of the ball.

2. How does a steel ball oscillate in a tube?

A steel ball oscillates in a tube due to the force of gravity pulling it down and the force of the tube pushing it back up. As the ball moves up and down, it creates a back and forth motion known as oscillation.

3. What factors affect the oscillation of a steel ball in a tube?

The oscillation of a steel ball in a tube can be affected by several factors, such as the mass of the ball, the amplitude of the oscillation, and the length and diameter of the tube. Other factors that can impact the oscillation include air resistance and the surface of the tube.

4. How does the amplitude of oscillation affect the motion of the steel ball?

The amplitude of oscillation, or the maximum displacement of the ball from its equilibrium position, directly affects the motion of the steel ball. A larger amplitude will result in a longer period of oscillation and a greater distance traveled by the ball.

5. What is the equation for the period of oscillation of a steel ball in a tube?

The equation for the period of oscillation of a steel ball in a tube is T = 2π√(m/k), where T is the period in seconds, m is the mass of the ball in kilograms, and k is the spring constant of the tube in newtons per meter.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
680
  • Introductory Physics Homework Help
Replies
9
Views
771
  • Introductory Physics Homework Help
Replies
4
Views
435
  • Introductory Physics Homework Help
Replies
18
Views
633
  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Introductory Physics Homework Help
2
Replies
38
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
289
  • Introductory Physics Homework Help
Replies
3
Views
939
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
4K
Back
Top