An aluminum clock pendulum having a period of 1.00 s keeps perfect time at 20 degrees celcius. (A) When placed in a room at a temperature of -5.0 Celcius, will it gain or lose time? (B)How much time will it gain or lose every hour.
this concerns change in length of aluminum due to fall in temp T(20 C) = 2pi√(L20/g = 1 sec L20 = g/4pi^2 ----------- (1) T(-5 C) = 2pi√(L5/g L5 = T(5)^2*g /4pi^2 ----------- (2) coefficient of linear expansion of aluminum = alpha alpha = delta L/L* delta T 23*10^-6 = (L5 - L20) /L20*(- 5 - 20) (L5 - L20) /L20 = - 25*23*10^-6 L5/L20 = 1 - 25*23*10^-6 = 0.999425 --- (3) L5 decrease in length T(5) will reduce (2) / (1) L5/L20 = [T(5)^2*g /4pi^2] * [4pi^2/g] = 0.999425 [T(5)^2] = 0.999425 T(5) = 0.9997 sec this is period when in (-5 C) room a) pendulum gains time in cold room word gains is used because it takes less in 1 oscillation. b) it gains time = 1 - 0.9997 = 0.0003 sec in 1 sec gain in every hour = 0.0003*3600 = 1.08 sec
A for a swinging pendulum period T =2pi*√(L/g), where L is length of the pendulum, g=9.8m/s^2; metals shrink when cooled, so period will decrease, number of ticks per hour will increase, the clock will hurry up; B? period at t1°=20C° is T1=2pi*√(L1/g); ? period at t2°=-5C° is T2=2pi*√(L2/g); ? the law of linear extension says L2/L1= 1+s*Δt°, where s=23.1·10-6 (1/К°) is specific linear extension of aluminum, Δt°=t2°-t1°; ? thus T2/T1 =√(L2/L1) = √(1+s*Δt°) =f1/f2, where frequency f1= 1 Hz at 20C° or 3600 ticks per hour, hence frequency at -5C° is f2= f1/√(1+s*Δt°); therefore instead of 3600 ticks per hour the clock will do 3600/√(1+s*Δt°) ticks per hour; thus it will gain 3600*(1/√(1+s*Δt°) –1) = = 3600*(1/√(1 -23.1·10-6 *25°) –1) = 1.04 s/hour;
