Answer:
Time period of a simple pendulum, \[T=2\pi \sqrt{\frac{l}{g}}\] \[\therefore \] \[g=\frac{4{{\pi }^{2}}l}{{{T}^{2}}}\] Clearly, the error in the measurement of time period T has larger effect on the value of g than the error in the measurement of length \[l\]. Reasons (i) T is very small. (ii) In contrast to \[l,={{T}^{2}}\] appears in the formula for g. To minimise the error, time period for a large number of oscillations is measured.
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