A grandfather pendulum has a simple brass rod of length L. One night, the temper
ID: 2206744 • Letter: A
Question
A grandfather pendulum has a simple brass rod of length L. One night, the temperature in the house is 24.0 oC and the period of the pendulum is 1 second. The clock keeps correct time at this temperature. 1) What is the length of the pendulum's rod? L = cm 2) The temperature in the house quickly drops to 16.0 oC just after 10PM, and stays at that value. By how much does the length of the pendulum change? The coefficient of linear expansion for brass is ? = 1.9E-05 K-1. ?L = x 10 -6 m 3) After the temperature drops to 16.0 oC at 10PM, it stays at that value. How much does the period of the pendulum change after the temperature has dropped to 16.0? ?T = x 10 -6 s 4) How many cycles does the pendulum make between 10 PM and the next morning when the pendulum indicates 10 AM? N =Explanation / Answer
T = 2 pi sqrt(L/g) gives the natural period of the pendulum. When the temperature drops, so does "L", and you need a thermal expansion coefficient for brass. Look for the one in your textbook, but I'm using the one from Wikipedia: 19 x 10^(-6) per degree C. This would reduce the length by [19 x 10^(-6) ](25-18.1) = 0.0001311 so that the new length will be 0.9998689 times the old length, and the new period will be 1 seconds times sqrt(0.9998689) = about 0.999934 seconds. The number of seconds between 10 pm and 10 am is 43200, so the clock has swung 43200 times when it reads 10:00 in the morning, but the actual time required for those swings is "short" by 43200 x 0.000066 seconds = 2.8 seconds. The actual time would be 9:59:57.2, and the clock is not noticeably wrong.