Brass Pendulum Bob
Physics help? The pendulum of a clock consists of a small heavy bob...?
The pendulum of a clock consists of a small heavy bob at the end of a brass rod. The clock keeps accurate time at 20°C, at which point the pendulum has a period of 0.5520s when the temperature is 35°C, will the clock be fast or slow? By how much will it be in error after one day at 35°C?
help?D: please explain in a LOT of detail>.<
By T = 2π x√[L/g]
=>As the L increases with temperature =>T will increase =>the clock will be slow
By Lt = Lo[1+ α x ∆t*C]
=>Lt = Lo[1 + 19 x 10^-6 x (35-20)]
=>Lt = 1.000285Lo
Thus T20*C = 2π x√[Lo/g] --------------(i)
& T35*C = 2π x√[Lt/g] ----------------(ii)
=>By (ii)/(i) :-
=>T35*C/T20*C = √[Lt/Lo]
=>[T35*C/T20*C] = √1.000285
=>T35*C = 0.5520 x 1.0001424898 = 0.5520786544 sec
Thus ∆t/sec = 0.5520786544 - 0.5520 = 7.86544 x 10^-5
Thus ∆t in 24 hour = 24 x 60 x 60 x 7.86544 x 10^-5
=>Thus ∆t in 24 hour = 6.80 sec