Question

Your pendulum clock which advances 1.0 s for every complete oscillation of the pendulum is running perfectly on Earth at a site where the magnitude of the acceleration due to gravity is 9.80 m/s2. You send the clock to a location on the Moon where the magnitude of the acceleration due to gravity is 1.65 m/s2.

Answer 1

As we know that time period of simple pendulum will be

$T = 2\pi \sqrt{\frac{L}{g}}$

now we will have

time period of a clock on the surface of earth will be 1 s where we have acceleration due to gravity is g = 9.8 m/s/s

now if we took the pendulum to the surface of moon where acceleration due to gravity will be 1.65 m/s/s then this time period will change

so we will say by above equation

$\frac{T_{earth}}{T_{moon}} =\sqrt{ \frac{g_{moon}}{g_{earth}}}$

now we will have

$T_{moon} = \sqrt{\frac{g_{earth}}{g_{moon}}}T_{earth}$

now plug in all data in this

$T_{moon} = \sqrt{\frac{9.8}{1.65}}(1s)$

$T_{moon} = 2.44 s$

so time period on moon will be 2.44 s