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The time period of a simple pendulum is given by the formula, `T = 2pi sqrt(l//g)`, where T = time period, l = length of pendulum and g = acceleration due to gravity. If the length of the pendulum is decreased to 1/4 of its initial value, then what happens to its frequency of oscillations ? |
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Answer» The time period of a simple pendulum is `T = 2pi sqrt(l//g)`. When length is decreased to `1//4^(th)` of its initial value then `l^(1) = l//4`. `therefore` New time period, `(T^(1))` `= 2pi sqrt((l//4)/(g)) = (2pi)/(2) sqrt(l/g) = 1/2 (T)` `therefore` Time taken to perform one oscillation is reduced to half of its initial value. `therefore` Frequency of oscillation becomes double that of the initial value. |
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