If a fixed length simple pendulum is found to have three times the period on an...
- If a fixed length simple pendulum is found to have three times the period on an unknown planet’s surface (compared to Earth), what is the acceleration due to gravity on that planet? Show your work.
Homework Answers
Time period of a simple Pendulum is given by:
T = 2*pi*sqrt (L/g)
Since Given that Length of pendulum is fixed, So from above equation Period is inversely proportional to the square root of acceleration due to the gravity on the planet. Which means
T2/T1 = sqrt (g1/g2)
Given that T2 = 3*T1
Where T1 = time period on earth
g1 = gravity on earth = 9.81 m/sec^2
So,
g2 = g1*(T1/T2)^2
g2 = 9.81*(1/3)^2
g2 = 1.09 m/sec^2 = acceleration due to gravity on that planet
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