Question

1) Suppose you want to use a simple pendulum to measure accurately the acceleration of gravity around the Earth. Gravity on the Earth's surface varies by around 0.7%, from 9.7639 m/s² on the Nevado Huascarán mountain in Peru to 9.8337 m/s² at the surface of the Arctic Ocean (according to Wikipedia). Your pendulum has a length of 0.750 m, and you time its oscillations while you count 200 full periods in each location. What is the maximum error, in seconds, that you can make in the time difference if you want to measure the difference in period between those two locations to 10% precision? (Note: because ? is proportional to the inverse square of the period, a 10% precision in a measurement of the period will only give a 20% precision in a measurement of ?. If you want to do better, you can count more oscillations, or you can measure the times more accurately, or both.)

Answer 1

max. error allowed in period difference = 10%

fractional error allowed in (period diff.) T = 0.1

Tn (Nevado)and Tp ( Peru) are the periods at the two locations

Tn = 2(l/g)^1/2 = 2 * sqrt(0.75/9.7639) =1.7414 s

Tp = 2 * sqrt(0.75/9.8337) =1.7352 s

  T = 1.7414 - 1.7352 = 0.0062 s

let error in each period be r sec. The error in

T = 2r

fractional error in T = 2r/0.0062

= 0.1 , allowed fractional error

r = 0.1*0.0062/2 = 0.00031 sec

maximum error allowed in period measurement = 0.00031 s