Ask your own question, for FREE!
Trigonometry 8 Online
OpenStudy (kailee503):

a simple pendulum has a period of 7 seconds on earth. What would the approximate period be on the moon, where the gravity is approximately 1/6 as strong as on earth

OpenStudy (danjs):

recall if you assume a small angle displacement the period of a pendulum is T depends on only the Length of pendulum L and the gravity force g \[T=2\pi*\sqrt{\frac{ L }{ g }}\]

OpenStudy (danjs):

period T=7 s, and earth gravity g=9.8 m/s^2 \[7 = 2\pi \sqrt{\frac{ L }{ 9.8 }}\] \[L \approx 12.16~m\] The pendulum has a length around 12.16 meters

OpenStudy (danjs):

1/6 earth gravity for the moon is g=9.8/6 m/s^2 now have the length L=12.16 , \[T=2\pi \sqrt{\frac{ 12.16 }{ (9.8/6) }}\approx 17.15~~s\]

OpenStudy (kailee503):

Thank you very much!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!