course phy232
I had, like, three or four other problems typed in, and for some unknown reason, I put my book up beside the keyboard and it all got erased. It wouldn't even let me undo it through the ctrl+ z command. It was just the pasta problem, and the heat at the junction of three bars in the shape of a Y. I may retype those up for resubmission or just show you in class.
17.124 Challenge problemA pendulum clock is designed to tick off one second on each side-to-side swing of the pendulum (two ticks per complete period). Will the pendulum clock gain time in hot weather and lose it in cold, or the reverse? Explain your reasoning.
To me, the pendulum will lose time as the temperature increases. My reasoning comes from the fact that throughout this chapter in thermodynamics we've constantly established the fact that things expand as the temperature increases. So if the hands of the clock become longer (not by much though), it seems perfectly plausible to believe that the longer the pendulum swings, the more time it will take, thus the clock will lose time.
B.) a particular pendulum clock keeps correct time at 20.0 Celsius. The pendulum shaft is steel and its mass can be ignored with that of the bob. What is the fractional change in length when it is cooled to 10.0 Celsius.
Maybe it would seem likely that we can use the fact that the change in length will be equal to the coefficient of linear expansion * change in temperature * initial length . Maybe we can represent this with variables. Maybe we can use numbers.
The temp change will be 10-20=-10K
beta= 1.2*10^-5
So we have d'L= initial length * 1.2*10^-4 or d'L=.00012*Initial Length
Does it make sense to say that .00012 is the fractional representation with which the length contracts?
Sure does; however you would use alpha for length expansion
c.) how many seconds per day will the clock gain or lose at 10.0 Celsius?
Provided this is right, the length of each swing is smaller so the the time will speed up in the clock. I think I may have to think back to simple harmonic motion?
If the normal length of the clock makes one period in one second, then length of contraction is going to slow it down by .00012 of this...there's 3600 seconds in a hour...86400 seconds in a twenty-four hour day... if you take this and multiply it by .00012 then you get 10.368..so the seconds in each day would increase by 10.368 seconds...
A fractional change of .00012 in length would result in a fractional change of .00006 is period, since period is proportional to the square root of length. .00006 * 84600 sec is about 5 sec, half of what you say.
T = k sqrt(L), where T is period and L is length.
Thus dT/dL = k / (2 sqrt(L)) so using the differential
`dT = k / (2 sqrt(L) ) * `dL.
If `dL = .00012 * L then `dT = k / (2 sqrt(L) ) * .00012 * L = .00012 k sqrt(L) / 2 = .00012 / 2 * k sqrt(L) = .00006 * T.
How closely must the temperature be maintained for the clock not to lose or gain a second in each day? Well, I'm sure this answer requires more than just an explanation, even though it says that the clock pendulum stays in working order at 20 Celsius.
A deviation of 10 degrees results in about a 5-sec difference. What deviation would result in a 1-sec difference?