A certain frictionless simple pendulum having a length L and mass M swings with

Question

A certain frictionless simple pendulum having a length L and mass M swings with period T. If both L and M are doubled, what is the new period? A) 4 T B) 2T C) 2T D) T E) T/4 A frictionless pendulum clock on the surface of the earth has a period of 1.00 s. On a distant planet, the length of the pendulum must be shortened slightly to have a period of 1.00 s. What is true about the acceleration due to gravity on the distant planet? A) The gravitational acceleration on the planet is slightly greater than g. B) The gravitational acceleration on the planet is slightly less than g. C) The gravitational acceleration on the planet is equal to g. D) We cannot tell because we do not know the mass of the pendulum.

Explanation / Answer

here,

initial the length of pendulam is l and the time period is T

T = 2 * pi * sqrt(l/g) ...(1)

and when l' = 2 * l

T' = 2 * pi * sqrt(2 * l/g)

T' = 2 * pi * sqrt(l/g) * sqrt(2)

T' = sqrt(2) * T

the new time period is sqrt(2) * l

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