A pipe in the shape of a right-angle elbow contains a vertical section and
a horizontal section with a valve at the right angle. Presently the valve
is closed and the vertical section of the pipe contains a column of water
having length L. When the valve is opened, the amount of water that
passes through the valve, as a function of time is given by
x(t) = L {1 − cos (sqrt(g/L)*t)}
for 0 ¡Ü x ¡Ü L, where g = 9.8 meters per seconds squared is the acceleration
due to gravity. If the initial column length is L = 2.0 meters
a.) how long (in seconds) does it take for one-half of the water column to flow
pass the valve?
b.) How long (in seconds) does it take all of the water column to flow pass the
valve?
some of the symbols did not appear correctly, so I am guessing.
The volume of water that has passed at time T is
Volume=int x(t)dt from 0 to T
one half the initial amount is 1/2 Area*Length
so
1/2 Area*2.0=INt 2.0(1-cos(sqrt g/2))t dt
hmmm. Area of the column is not given.So I wonder about the units of x(t). Seeing on the right, it is L, so I guess the problem maker assumed unit area
1=2 int dt -2int cos(sqrt g/2) t)dt
1=2T+2sin (sqrtg/2)T
it is a quadratic in T, so use the quadratic equation to find T. check my thinking, I did most of this ASCII typing and figuring in my head.