What will happen to time period of pendulum when taken at depth of half radius of earth pls tell me and dont copy ×××
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The acceleration due to gravity on the earth’s surface averages to 9.8 m/s^2 while on the moon it averages to 1.63 m/s^2. The formula for determining the period of the pendulum is shown below.

According to the formula the period of the pendulum T is inversely proportional to the square root of of g and is directly proportional to the square root of L. That implies that as the value of g is lowered its period will increase.
Since the value of g on the moon is 6 times smaller than that of the earth its period if not corrected becomes longer by the square root of 6. If it beats two second per full swing on earth it beats 4.90 seconds per full swing on the surface of the moon.
To correct that longer period its length L must be made shorter by the square root of 6. If it is 99.2 cm on earth it must be shortened to 40.5 cm on the surface of the moon.
t=2pie√l/9

According to the formula the period of the pendulum T is inversely proportional to the square root of of g and is directly proportional to the square root of L. That implies that as the value of g is lowered its period will increase.
Since the value of g on the moon is 6 times smaller than that of the earth its period if not corrected becomes longer by the square root of 6. If it beats two second per full swing on earth it beats 4.90 seconds per full swing on the surface of the moon.
To correct that longer period its length L must be made shorter by the square root of 6. If it is 99.2 cm on earth it must be shortened to 40.5 cm on the surface of the moon.
t=2pie√l/9
Soumya5887:
Hey i said in the question that solve my question and u have given wrt moon
sry
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