A pendulum clock with a brass suspension system is calibrated so that its period is 1 s at 20?C. If the temperature increases to 41 ?C, by how much does the period change? Answer in units of s.I tried this twice and got 0.000193162s and 0.01965s, both wrong. Any help explaining the problem? Thanks
T 2π (L/g)^1/2 dT/dL π/L (L/g)^1/2 dT π/L (L/g)^1/2 dL where dL L λ dTemp then dT π (L/g)^1/2 λ dTemp 1/2 T λ dTemp λ 1,9x10^-5 k^1 then dT 0.5*1*1.9x10^-5*21 1.995x10^-4 s 0.0001995
