The basic procedure for this problem is not too hard but get

The basic procedure for this problem is not too hard, but getting details of the calculation correct is NOT easy, and may take some time. A ping-pong ball is caught in a vertical plexiglass column in which the air flow atternates sinusoidally with a period of 60 seconds. The air flow starts with a maximum upward flow at the rate of 5.1m/s and at t = 60 seconds the flow has a minimum (upward) flow of rate of - 9.1 m/s. (To make this clear: a flow of - 5m/s upward is the same as a flow downward of 5m/s. The ping-pong ball is subjected to the forces of gravity (-mg) where g = 9.8m/s^2 and force due to air resistance which is equal to k times the relative velocity of the ball through the air in MKS units.

Solution

Average velocity Vmean = (Vmax + Vmin)/2 = (5.1 - 9.1)/2 = -2 m/s

Amplitude A = Vmax - Vmean = 5.1 - (-2) = 7.1 m/s

Frequency w = 2*pi/T = 2*pi/60 rad/s

Since we know that cos function has max at t= 0 and min at t = T/2, the function can be of the form

A(t) = 7.1*cos(2*pi/60 *t) - 2

Force due to air resistance = -k*(v - A(t))

= -k*[v - 7.1*cos(2*pi/60 *t) + 2]

Net force on ball F = -mg -k*[v - 7.1*cos(2*pi/60 *t) + 2]

F = ma = m*dv/dt

m*dv/dt = -mg -k*[v - 7.1*cos(2*pi/60 *t) + 2]

dv/dt = -g -(k/m)*[v - 7.1*cos(2*pi/60 *t) + 2]

dv/dt = -9.81 -(k/m)*[v - 7.1*cos(2*pi/60 *t) + 2]

For complimentary solution, we write v\' = -(k/m)*v

V_comp = C*e^(-kt/m)

For a particular solution, let V_p = A*sin(2*pi/60*t) + Bt

dv/dt= A*2*pi/60 *cos(2*pi/60*t) + B

A*2*pi/60 *cos(2*pi/60*t) + B = 7.1*k/m*cos(2*pi/60 *t) - 2k/m - 9.81

Comparing, A = 7.1*k/m*60/(2*pi), B = -(2k/m + 9.81)

Hence V_p = 7.1*k/m*60/(2*pi) *sin (2*pi*t/60) - (2k/m + 9.81)*t

Adding the complementary and particular solutions we get,

V(t) = 7.1*k/m*60/(2*pi) *sin (2*pi*t/60) - (2k/m + 9.81)*t + C*e^(-kt/m)

We get V(0) = C

1.4 = C

V(t) =  7.1*k/m*60/(2*pi) *sin (2*pi*t/60) - (2k/m + 9.81)*t + 1.4*e^(-kt/m)