Let x1(t) and x2(t) be functions that represent the horizontal displacement of two simple pendulums from their central position at time t (in seconds)
x1(t) = 0.5sin(pi(t))/5)
x2(t) = 0.1sin (pi(t))/4)
If both pendulums start at the same displacement at time t = 0, at what time are they both next at the same displacement?
You would be solving:
.5sin(?t/5) = .1sin(?t/4)
5sin(?t/5) = sin(?t/4)
getting a "feel" for the question, I ran it through Wolfram, had to change the t to x
http://www.wolframalpha.com/input/?i=plot+y+%3D++.5sin(%CF%80x%2F5)+,+y+%3D+.1sin+(%CF%80x%2F4)+from+0+to+3%CF%80
looks like about x or t = appr 5.3
let y = 5sin(?x/5) - sin(?x/4)
using Newton's method:
dy/dx = ?cos(?x/5) - (?/4)cos(?x/4)
newx = x -(5sin(?x/5) - sin(?x/4) / ( ?cos(?x/5) - (?/4)cos(?x/4) )
using a starting x of 5.3
newx = 5.2685..
newx = 5.2685 , correct to 4 decimals
check:
x1(5.2685) = .5(sin 3.310...) = .0839...
x2(5.2685) = .1sin(4.1379..) = .0839..
not bad!
To find the time at which the two pendulums are next at the same displacement, we need to solve the equation:
x1(t) = x2(t)
Substituting the given functions for x1(t) and x2(t), we have:
0.5sin(pi(t))/5) = 0.1sin (pi(t))/4)
To simplify the equation, we can cross multiply and get rid of the denominator:
(0.5sin(pi(t))(4)) = (0.1sin(pi(t))(5))
Now we can cancel out the common factors:
2sin(pi(t)) = sin(pi(t))
Next, we simplify the equation by using the trigonometric identity:
sin(a) = sin(b) if and only if a = b or a = pi - b
So we have two possibilities:
1) pi(t) = pi(t)
2) pi(t) = pi - pi(t)
For the first possibility, we can see that it is true for all values of t.
For the second possibility, we can solve for t:
pi(t) = pi - pi(t)
2pi(t) = pi
t = 1/2
Therefore, the two pendulums will be next at the same displacement at t = 1/2 seconds.