Maybe you could use three sine waves with differing periods?
Hey guys, got given this question by my teacher and I'm not really sure where to go with it.
"A clock starts at exactly 12:00. At what times over the span of 12 hours do all three hands on the clock line up?"
-assuming all hands move at a constant velocity, rather then start and stop every second
My first thought is some sort of simultaneous equations, but I haven't been able to get very far with it.
Thanks
Sent from my SM-G935F using Tapatalk
HSC 2018: 3U | 4U | Chemistry | Physics | Advanced English |
Maybe you could use three sine waves with differing periods?
Last edited by fan96; 4 Feb 2018 at 5:59 PM.
HSC 2018: English Adv. [80] • Maths Ext. 1 [98] • Maths Ext. 2 [95] • Chemistry [87] • Software Design [95]
ATAR: 97.40 | Uni Course: B Advanced Mathematics (Hons) / B Engineering (Hons) (Computer) at UNSW
Assuming a circular clock with hands undergoing uniform circulation motion, the speeds of the hands in revolutions per hour (rph) are:
• second hand: 60 rph (since it undergoes one revolution per minute);
• minute hand: 1 rph;
• hour hand: 1/12 rph (since it takes 12 hours for it to undergo a full revolution).
Hence the relative speed of the
• second hand to minute hand is 59 rph;
• minute hand to hour hand is 11/12 rph.
Hence the
• second hand aligns with the minute hand every 1/59 hours (call this event A);
• minute hand aligns with the hour hand every 12/11 hours (call this event B).
The three hands align precisely when event A and event B are in occurrence simultaneously. So we are looking for the first time event A and event B happen together. Note that an obvious time when they occur together is when it's a twelve o'clock again, so after 12 hours. Can it happen any sooner? It turns out no. An explanation of this is below.
Suppose the events happen together for the first time at the m-th occurrence of event A and the n-th occurrence of event B, where m and n are positive integers and at 12:00, this represents the zeroth occurrence of each event. Then by comparison of time elapsed, we have
(1/59)*m = (12/11)*n.
So m/n = (12*59)/11.
Note the obvious solution of n = 11 and m = 12*59 corresponds to the obvious crossing time of the next twelve o'clock. The reason why there is no smaller solution for positive integer n (and hence no earlier crossing time) is that 11 is clearly coprime to 12*59 (they share no common factors other than 1), so the fraction (12*59)/11 is in lowest terms already.
In summary, the three hands all align only at the next twelve o'clock.
Last edited by InteGrand; 4 Feb 2018 at 6:40 PM.
There are currently 1 users browsing this thread. (0 members and 1 guests)
Bookmarks