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The time period of simple pendulum at the surface of earth is T. If it is taken to a height equal to the radius of earth. Find the period of new oscillations. |
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Answer» SOLUTION :At a height R. new `G^(1)= (GM)/(r^(2))= (GM)/(2R)^(2)=(g)/(4):. T PROP (1)/(sqrt(g))` `(T^(1))/(T_(0))= sqrt((g)/(g_(1)))= sqrt(((g)/(g))/(4))= 2`or `T^(1)= 2T_(0)` |
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