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A seconds pendulum is taken to planet where the acceleration due to gravity is 4 times that in earth .What would be the time period of the pendulum on the planet
Time period of a simple pendulum=2*pi*(l/g)1/2where l is the length of the string and g is the gravition of the planetSince it is a seconds pendulum , 2*pi*(l/g)1/2=1 second (in earth)On the other planet , gravity = 4gTherefore new time period= 2*pi*(l/4g)1/2 =2*pi*(l/g)1/2*(1/4)1/2=1 second * ½ =0.5 seconds
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