OpenStax College Physics for AP® Courses, Chapter 16, Problem 34 (Problems & Exercises)

This is College Physics Answers with Shaun Dychko. A clock gains 5.00 seconds per day and we are asked by what fraction should the pendulum length be changed such that it takes exactly 24 hours to indicate a day passing? So the time to indicate a day passing... in other words, the hour hand goes around the whole clock twice is 24 hours minus the 5.00 seconds that it's too early. So on this clock as it is to begin with, the hour hand will reach the very top 5 seconds before it should. So the actual amount of time that has passed is 24 hours minus this 5 second earliness. So this time is 24 hours times 3600 seconds per hour minus 5.00 seconds which is 86395 seconds. So what we are going to find is how many cycles of the pendulum is needed— so the pendulum is swinging below here— how many cycles are needed in order to indicate a full day passing? That will be the frequency of the pendulum multiplied by the time that it takes to indicate a day and the frequency is the reciprocal of the period and we make that substitution since we have a formula for period. So we have reciprocal of period one when we have the first length one times the time it takes for the whole day to be indicated and so that T 1 is 2π times square root L 1 over acceleration due to gravity so we take the reciprocal of that and substitute it in here so that's 1 over 2π times square root g over L 1 all times t 1— this is the number of cycles to indicate a full day passing— and that will be the same number of cycles after we change the length. So the thing that's constant here regardless of the length is the number of times that this pendulum has to swing back and forth. You know, each swing rotates the gear mechanism inside here by the same amount regardless of how long the pendulum is and so there's a certain number of swings necessary to turn the gears enough time such that a whole day is indicated to have passed. And so after adjusting the length, we'll take that same number of cycles there's no subscript 1 or 2 here— that's not necessary— because it's the same number of cycles in both cases but you know this number of cycles will take a different amount of time in the two cases. In the second case, the amount of time will be exactly 24 hours or converted into seconds, 86400 seconds, because in this after adjusting the length, this L 2 is meant to be a length such that the time will be exactly 1 day. So we are substituting with L 2 and T 2 here to find the same number of cycles in that means we can equate this and this because they are both equal to the number of cycles it takes to indicate a day. So we have 1 over 2π times square root g over L 1 times t 1 equals 1 over 2π times square root g over L 2 times t 2 and multiply both sides by 2π over square root g and then we get a whole bunch of things canceling here and we are also multiplying both sides by square root L 2 and we get square root L 2 over L 1 is t 2 over t 1 after you also divide both sides by t 1 so lots of things happening there. So we end up this line here: square root L 2 over L 1 is t 2 divided by t 1 and then square both sides and you get L 2 over L 1 is the square of t 2 over t 1 and then multiply both sides by L 1 and we get an expression for L 2 so the L 2 is gonna be t 2 over t 1 squared times L 1. So that's 86400 seconds—exact amount of time in a day— divided by 86395 seconds—which was the amount of time required to indicate a day with the incorrect L 1 length— and we square that and we get 1.0001158L 1 is L 2. So this means the length should be increased by 0.0116 percent if you take the difference and change it into a percent.