I really should know this, but I've been stumped on it since yesterday.


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

cant help ya out ashwin

..Okay, I got a few questions first..

1) Are we looking for the times or just the interval between when the second-hand and the minute-hand are superimposed?

2) Are we given a domain? Is this in the span of 24Hrs, 12Hrs, 1Hr or what? Based on what you have for your value equations for ω, it looks like a domain of an hour..

~Hmm

*scratches head* maybe this will help

www.physicsforums.com/showthread.php?t=150596

if you can do an essay on the Romantics for me...ill ask my dad for the answer to this. and it'll be the right answer...:)

..Okay, I got a few questions first..

1) Are we looking for the times or just the interval between when the second-hand and the minute-hand are superimposed?

2) Are we given a domain? Is this in the span of 24Hrs, 12Hrs, 1Hr or what? Based on what you have for your value equations for ω, it looks like a domain of an hour..

~Hmm
1) The time between each interval between the minute and second hand, which would occur after the second hand has rotated just past 2π radians.

2) The assumption is that the time of coincidence will be a constant throughout all coincidences of the minute and second hands, thus it makes no difference if it's in a one hour period, or a 3 million hour period.

I was given the answer after this was due, but it's still a little hazy for me. So far I understand:

At the angle of coincidence, the angle of the second hand (denoted as θ) will share the same reference angle of the minute hand (θ'):
θ = θ' + 2π


...Wait, nevermind, it just made the world of sense to me. Before I was trying to solve it based on them having the same angular speed, which makes no sense.

Both have to be solved for ω individually, but the two equations have to be equal such as the following:
Where θ=ωt and θ'=ω't,
ωt = ω't + 2π because of θ = θ' + 2π already being established

Therefore:
t(ω-ω') = 2π
t = 2π/(ω-ω')
= 2π/[(2π/60 seconds)-(2π/3600 seconds)]
= 3600/59 seconds....just over a minute

What the fuck?

Both have to be solved for ω individually, but the two equations have to be equal such as the following:
Where θ=ωt and θ'=ω't,
ωt = ω't + 2π because of θ = θ' + 2π already being established
..So basically we had to use Derivatives to solve this, but I don't see how you assumed the same Angular Velocity. I believe you defined the A.S. formula for the second- and minute-hand accurately. I didn't use them to start it, though, because I worked in units of seconds in the domain of an hour. Different method; same destination..

..But...I see said the blind man..


~Interesting

Remind me NOT to take Physics..

three fitty

fuck math, take arts

Alternatively, I think it's easier to find the angular velocity of the second hand relative to the minute hand. Just pi/30 - pi/1800

Then angular velocity equals theta/t

set theta equal to 2pi and solve for t

that's the consecutive time between the hands overlapping.

I know someone that will not the answer, called Smarties, and only Smarties has the answer.