The acceleration due to gravity on the surface of the moon is $$1.7m/s2$$. What is the time period of a simple pendulum on the surface of moon if it’s time period on the surface earth is $$3.5$$ seconds?(g$$ on the surface of earth is $$9.8m/s2).

Question

The acceleration due to gravity on the surface of the moon is $$1.7m/s2$$. What is the time period of a simple pendulum on the surface of moon if it’s time period on the surface earth is $$3.5$$ seconds?(g$$ on the surface of earth is $$9.8m/s2).

Infinity Learn · Accepted Answer

Given $$gm=1.7ms-2$$ $$ge=9.8ms-2Time$$ period on $$Earth=Te=3.5seconds$$                                    As $$Te=2$$$$π$$lge Time period on $$moon=Tm=2$$$$π$$lgm ratio of above two equations is,                                            $$TmTe=gegm$$                                              $$Tm=Tegegm$$                                               $$Tm=3.59.81.7=8.4$$ seconds

The acceleration due to gravity on the surface of the moon is $1.7 m / s^{2}$ . What is the time period of a simple pendulum on the surface of moon if it’s time period on the surface earth is 3.5 seconds?(g on the surface of earth is $9.8 m / s^{2}$ ).

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The acceleration due to gravity on the surface of the moon is 1.7m/s2. What is the time period of a simple pendulum on the surface of moon if it’s time period on the surface earth is 3.5 seconds?(g on the surface of earth is 9.8m/s2).

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

The acceleration due to gravity on the surface of the moon is $1.7 m / s^{2}$ . What is the time period of a simple pendulum on the surface of moon if it’s time period on the surface earth is 3.5 seconds?(g on the surface of earth is $9.8 m / s^{2}$ ).