Help I know I\'m close! A pendulum clock has an aluminum pendulum exactly 1.0m l

Question

Help I know I'm close! A pendulum clock has an aluminum pendulum exactly 1.0m long when the clock is calibrated perfectly. By how much is it off at the end of one day (seconds/day) if the temperature increases by 12 degrees C? Assume a simple pendulum in which the aluminum rod's mass is negligible compared to the pendulum bob. The coefficient of linear expansion for aluminum is 2.4*10^-5 The change in length I have calculated to the coefficient times the change in temperature = 2.88E-4=change in length Then I calculate t=square root of 9.8 and subtract t'= square of 9.8/change in length plus 1 and divide all that by seconds in a day but my answer is way off. Where am I going wrong??? 5 stars up for grabs

Explanation / Answer

T=sqrt(l/g)=sqrt(1/g) initially. T'=sqrt(l'/g)=sqrt((1+dl)/g) finally. You calculated dl=2.88e-4 m. So its period increases slightly...in a real day it will count fewer seconds by its tocks. T'/T=sqrt(1+dl)= 1.000144 So for 86400 s/day, it will count only 86387.56 s/day. Which is a difference of close to 12.5 seconds.

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.