Precalculus/Trig Question?

Question

The name of the game is to find the time between consecutive instances when the second hand and the minute hand of a clock coincide. This has to be done in relation to angular speed.

I've defined the angular speed of the second hand as:
ω = 2π/60 = π/30

Angular speed of the minute hand has been defined as:
ω = 2π/3600 = π/1800

Answer 1

ω1 = angular velocity of second hand in radians / sec
ω 2 = angular velocity of minute hand in radians / sec

ω1 = 2π / 60 radians / sec
ω2 = 2π / 3600 = π / 1800 radians / sec

RELATIVE angular velocity is given by:-
π / 30 - π / 1800 radians / sec
= (59 π / 1800) radians / sec

59π / 1800 radians<-----> 1 sec
2π radians<--->2π / (59π / 1800) sec
= 3600 / 59 sec
= 61 sec

Will coincide after 61 sec.

Answer 2

So what you are really doing is finding times that they will coincide again. Since they run at different rates, it will take more than one complete revolution for them to meet up again. Essentially the second hand will spin all the way around the clock and then catch up to the minute hand. So you want a time such that displaced angular distance of the second hand is 2pi + a and displaced angular distance of the minute hand is just a. You can draw a picture to see if this is right.

so t * pi/30 = 2pi +a
and t* pi/1800 = a
giving a system of 2 equations with 2 unknowns

so t * pi/30 = 2pi + t* pi/1800

so t*pi (60/1800 - 1/1800) = 2pi (can cancel the pis out)

t * (59/1800) = 2

t = 61.02 or 3600/59 seconds

Intuitively, as one would expect it is just over 1 minute. After the first 60 seconds the minute hand has moved forward 6 degrees and the second hand has made a complete revolution. Now in the next second the second hand moves forward 6 degrees and the minute hand then moves forward an infinitesimal amount.