Piston Problem
A piston moves a distance of 12cm from top to bottom (starts at top)
One complete piston movement takes 0.04 seconds.
The equation for the distance d = 6Cos(50πt)
where t = time in seconds
And d is the vertical distance from point ‘P’ (cm)
During the first cycle how
Distance
long is the piston less than
‘d’
2cm above point ‘P’?
Point P
(Where point ‘P’ is the
central position of the
vertical movement)
Method A
Method B
Method C
Review
Piston Problem method A
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A piston moves a distance of 12cm from top to bottom (starts at top)
One complete piston movement takes 0.04 seconds.
Let A = 50πt
Solve 2 = 6Cos(50πt)
The equation for the distance d = 6Cos(50πt)
where t = time in seconds
becomes 2 = 6Cos(A)
And d is the vertical distance from point ‘P’ (cm)
0.333 = Cos(A)
During the first cycle how
-1 0.333 = 1.23
Distance
A
=
Cos
long is the piston less than
‘d’
Other solution for A
2cm above point ‘P’?
Point P
OR A = 2π – 1.23 = 5.05
(Where point ‘P’ is the
Use A to find ‘t’
central position of the
vertical movement)
As ‘t’ = A ÷ 50π
6
4
2
– 2
– 4
– 6
d
t = 0.007837
t = 0.03216
0.01 0.02 0.03
t
time difference
0.03216 – 0.007837 = 0.0243 sec
Piston Problem method B
Home
A piston moves a distance of 12cm from top to bottom (starts at top)
One complete piston movement takes 0.04 seconds.
The equation for the distance d = 6Cos(50πt)
Solve 2 = 6Cos(50πt)
where t = time in seconds
And d is the vertical distance from point ‘P’ (cm)
0.333 = Cos(50πt)
During the first cycle how
-1 0.333
Distance
(50πt)
=
Cos
long is the piston less than
‘d’
2cm above point ‘P’?
50πt = 1.23
Point P
t = 0.007837
(Where point ‘P’ is the
central position of the
OR 50πt = 5.05
vertical movement)
6
4
2
– 2
– 4
– 6
d
t = 0.03216
0.01 0.02 0.03
t
time difference
0.03216 – 0.007837 =
t = 0.0243 sec
Piston Problem method C
Home
A piston moves a distance of 12cm from top to bottom (starts at top)
One complete piston movement takes 0.04 seconds.
Solve 2 = 6Cos(50πt)
The equation for the distance d = 6Cos(50πt)
where t = time in seconds
0.333 = Cos(50πt)
And d is the vertical distance from point ‘P’ (cm)
(50πt) = Cos-1 0.333
During the first cycle how
Distance
long is the piston less than
50πt = 1.23
‘d’
2cm above point ‘P’?
t = 0.007837
Point P
(Where point ‘P’ is the
central position of the
vertical movement)
6
4
2
– 2
– 4
– 6
Length of 1cycle?
50πt cycles = 2π
1 cycle = 2π ÷ 50π = 0.04
2nd time
0.04 – 0.007837 = 0.03216
d
0.01 0.02 0.03
t
0.04
time difference
0.04 – 0.03216 = 0.0243 sec
Piston Problem Review
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Solve 2 = 6Cos(50πt)
Let A = 50πt
Simplify the equation 2 = 6Cos(A)
Rearrange to give ? = Cos(A)
Solve ? = Cos(A)
Find other solutions for Solve A = ?
Use each ‘A’ solution to
find a value for t = ?
Answer Q time difference = ?
6
4
2
– 2
– 4
– 6
d
Solve 2 = 6Cos(50πt)
becomes 2 = 6Cos(A)
0.333 = Cos(A)
A = Cos-1 0.333 = 1.23
Other solution for A
OR A = 2π – 1.23 = 5.05
Use A to find ‘t’
As ‘t’ = A ÷ 50π
t = 0.007837
t = 0.03216
0.01 0.02 0.03
t
time difference
0.03216 – 0.007837 = 0.0243 sec