High School Physics The problem of the hour, minute, and second hands overlapping

At 0:00:00, the first time it coincided.

The angular velocity of the hour hand is 360 degrees per 12 hours, i.e. 1 120 degrees per second.

The angular velocity of the minute hand is 360 degrees every 60 minutes, i.e. degrees per second.

The angular velocity of the second hand is 360 degrees every 60 seconds, that is, 6 degrees per second.

There is a second coincidence after an interval of x seconds.

1 120*x+360n=formula.

Equation (n, k are integers).

By 1 formula: n=11x (3600*12).

By 2 formula: k = 59 x 3600

Therefore, in order to make n and k integers, the minimum of x should be 3600*12, that is, x=12*3600 seconds=12 hours.

This question can be understood as a problem of chasing in circles).

If the angular velocity of the hour hand is w, the angular velocity of the minute hand is 60w, and the angular velocity of the second hand is 3600w, then the three hands must have a second hand that is faster than the minute hand n turns (i.e. 2n) and the minute hand faster than the hour hand m turns (2m).

3600wt-60wt=2nπ 60wt-wt=2mπ （2π/w=t=1h)

The clock is 12 hours 360 degrees, so it's 1 120 per second.

The minutes are 60 minutes 360 degrees, so it's 1 10 per second.

Seconds are 60 seconds, 360 degrees, so it's 6 degrees per second.

Coincident after x seconds.

Then: 1 120*x+360h=1 10*x1 10*x+360t=6*x (h, t is an integer) is solved: x=12 hours.

I think it's 22 times, any time you start counting.

If it is simplified to 60 steps of the second hand and 1 step of the minute hand, and the minute hand 60 steps and the hour hand 1 square (5 steps), that is, the minute hand is 12 steps and the hour hand is counted in one step, the time that coincides from 0 should be: 0:0:

Illustrates a problem, every hour will coincide times and the position is getting closer and closer to the number of the next hour, and then the specific point is, starting with 0:0:0, from this time onwards, there is no overlap from 0 to 1 o'clock, and there is a coincidence within 1 o'clock to 2 o'clock, and the position is 1:

At 5:5, the clock has shifted a little to 2 o'clock, and the coincidences between 2 and 3 o'clock are closer to 2 o'clock than last time, until this situation: 11 o'clock to 12 o'clock, the position is the limit, and it completely coincides to 12:

0:0, at this time, the hour hand went around once, a total of 11 times, the hour hand walked twice a day, 22 times, if there is anything wrong, please advise.

2 times I will encounter this problem when playing the game every day.

The rotation of the second hand is very fast, so you can ignore it, mainly depending on the number of times the hour and minute hands coincide.

The minute and hour hands coincide a little more than once every hour, around 1:05, 2:10, 3:15, and again after 11:55. This makes a total of 11 times. There was one more time at the last 12 o'clock, 12 times.

If you add the first 12 o'clock, there are 13 times.

Calculation Details:

The angular velocity of the hour hand: 2 * 12 per hour i.e. 2 * 12 radians per hour minute hand: 2 per hour i.e. one turn per hour.

The time when they meet once is t

Angular velocity of the minute hand t) - (angular velocity of the hour hand at t) = 2

If you count the first one, it should be 12 times.