Forced Vibrations: ping-pong ball in vertical column

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jjm5119

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A ping-pong ball is caught in a vertical plexi-glass column in which the air flow alternates sinusoidally with a period of 60 seconds. The air flow starts with a maximum upward flow at the rate of -0.4 m/s and at t = 30 seconds the flow has a minimum (upward) flow of rate of -0.4 m/s. (To make this clear: a flow of -5m/s upward is the same as a flow downward of 5 m/s.

The ping-pong ball is subjected to the forces of gravity (-mg) where g= 9.8m/s^2 and forces due to air resistance which are equal to k times the apparent velocity of the ball through the air.

a: What is the average velocity of the air flow?

b: Write a formula for the velocity of the air flow as a function of time.

c: Write the differential equation satisfied by the velocity of the ping-pong ball (relative to the fixed frame of the plexiglass tube.) The formulas should not have units entered, but use units to trouble shoot your answers. Your answer can include the parameters m, k, t, and v

d: Find the amplitude and phase shift of this solution. You do not need to enter units.

e: Find the general solution by adding on a solution to the homogeneous equation. Calculate the specific solution that has initial conditions t=0 and w(0) = 4
 
jjm5119 said:
A ping-pong ball is caught in a vertical plexi-glass column in which the air flow alternates sinusoidally with a period of 60 seconds. The air flow starts with a maximum upward flow at the rate of -0.4 m/s and at t = 30 seconds the flow has a minimum (upward) flow of rate of -0.4 m/s. (To make this clear: a flow of -5m/s upward is the same as a flow downward of 5 m/s.

The ping-pong ball is subjected to the forces of gravity (-mg) where g= 9.8m/s^2 and forces due to air resistance which are equal to k times the apparent velocity of the ball through the air.

a: What is the average velocity of the air flow?

b: Write a formula for the velocity of the air flow as a function of time.

c: Write the differential equation satisfied by the velocity of the ping-pong ball (relative to the fixed frame of the plexiglass tube.) The formulas should not have units entered, but use units to trouble shoot your answers. Your answer can include the parameters m, k, t, and v

d: Find the amplitude and phase shift of this solution. You do not need to enter units.

e: Find the general solution by adding on a solution to the homogeneous equation. Calculate the specific solution that has initial conditions t=0 and w(0) = 4
Click to expand...

This is a standard 2nd. order ODE problem.

Start writing Newtons equation(F= ma) - and solve.

Please show your work/thought so that we know where to begin to help you. If you are in an engineering curriculum - you better learn how to tackle these problems.
 
subhotosh khan

A. What is the average velocity of the air flow?

Vavg = (-.4 + (-.4))/2 = -.4 m/s

B. Write a formula for the velocity of the air flow as a function of time.

A(t) = A + Bsin((pi)t/30) then use initial conditions of A(0) = -.4 and A(30) = -.4 to solve for the coefficients of A and B.


Now where would you go from there for parts c, d, and e?
 
bwat687 said:
subhotosh khan

A. What is the average velocity of the air flow?

Vavg = (-.4 + (-.4))/2 = -.4 m/s<<< airflow is following sine function - not a constant. You need to integrate overtime for average velocity.

B. Write a formula for the velocity of the air flow as a function of time.

A(t) = A + Bsin((pi)t/30) then use initial conditions of A(0) = -.4 and A(30) = -.4 to solve for the coefficients of A and B.


Now where would you go from there for parts c, d, and e?
Click to expand...

for part c - write f = ma to get the differential equation. You should get a second order ODE.

solve it to get d and e.
 
jjm5119 said:
A ping-pong ball is caught in a vertical plexi-glass column in which the air flow alternates sinusoidally with a period of 60 seconds. The air flow starts with a maximum upward flow at the rate of -0.4 m/s and at t = 30 seconds the flow has a minimum (upward) flow of rate of -0.4 m/s. (To make this clear: a flow of -5m/s upward is the same as a flow downward of 5 m/s.

Are these signs correct - how is it that maximum and minimum velocity has same magnitude and same direction (upward negative that is downward 0.4m/s - for both)


d: Find the amplitude and phase shift of this solution. You do not need to enter units.

e: Find the general solution by adding on a solution to the homogeneous equation. Calculate the specific solution that has initial conditions t=0 and w(0) = 4
Click to expand...
 
Yes i thought that was weird too, but that is how the problem is asked.
 
Im not really sure what you mean by the F = ma.....

Fg = force due to gravity = -9.8m (m/s^2)

Fair = k(A(t) - V) = K(B + Acos((pi)t/30))

F = ma
-9.8m (m/s^2) + K(B + Acos((pi)t/30)) = ma

-9.8m (m/s^2) + K(B + Acos((pi)t/30)) = mv'

my'' + Ky' = -9.8m (m/s^2) + K(B + Acos((pi)t/30))

Is this what you are referring to or not?
And if it is would that mean that v' = -9.8 (m/s^2) + (K/m)(B + Acos((pi)t/30))
Also for the second order ODE what would be the form of the particular solution?
Thanks for the help
 
NEVERMIND, SOLVED PROBLEM

ok so a. a= -.4

part b i get A=-.4 and B=0, there for A(t)= -.4

for part c i use f=ma and then i get ma=-9.8 +k(-.4)
then i use mv'+kv=-9.8m+k(-.4)
therefore the answer to c is v'=(-9.8m-.4k-kv)/m

i solve for my particular solution and it is -.4-(mg/k)

Now my amplitude will be zero and i can't figure out my phase shift. I have it as cos((pi/30)t- "something") I can't figure out the "something". I know pi/30 is correct and I know that my amplitude is correct. How do i determine the second part of the phase shift

thanks
 
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