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Unformatted text preview: Solutions for Homework 4 1 Solution 1: The hour hand is a uniform oscillator with period T = 1 = 60 minutes. The minute hand is a uniform oscillator with period T = 1 minute. The two oscillators are described by 1 = 2 60 and 2 = 2 if time is measured in minutes. The phase difference satisfies d dt ( 2 1 ) = 2 parenleftbigg 1 1 60 parenrightbigg = 2 59 60 Therefore the phase difference is itself a uniform oscillator, with period 60 59 . Thus the two hands are aligned again after 60 / 59 minutes, or 3600 / 59 61 . 0169 seconds. Solution 2: After one minute, the minute hand returns to where it started, but the hour hand has advanced by 1/60 of a rotation. The minute hand needs 1 / 60 of a minute, that is, one second, to get there. But in that time, the hour hand has advanced 1/60 of 1/60 of a rotation. The minute hand needs 1 / 60 of 1 / 60 of a minute to get there. And so on. So the time it takes for the minute hand to reach the hour hand is...
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This note was uploaded on 02/28/2011 for the course MATH 150 taught by Professor Dr.borgers during the Spring '09 term at Tufts.
- Spring '09