question_answer 15)

The acceleration due to gravity on the surface of the moon is \[\text{T=2 }\!\!\pi\!\!\text{ }\sqrt{\frac{\text{m}}{\text{k}}}\]. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth \[\therefore \])

Answer:

Here, \[\text{T=2 }\!\!\pi\!\!\text{ }\sqrt{\frac{\left( \text{m/2} \right)}{\text{k}}}\text{=2 }\!\!\pi\!\!\text{ }\sqrt{\frac{\text{m}}{\text{2k}}\text{.}}\] \[1\cdot 0\] \[\text{a=}\frac{\text{1}}{\text{2}}\text{m; }\!\!\omega\!\!\text{ =200 rev/min;}\] As, \[{{\text{ }\!\!\upsilon\!\!\text{ }}_{\text{max}}}\text{=a }\!\!\omega\!\!\text{ =}\frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ 200=100m/min}\text{.}\] \[1\cdot 7m{{s}^{-2}}\]\[3\cdot 5\] or \[=9\cdot 8m{{s}^{-2}}\]\[{{g}_{m}}=1\cdot 7m{{s}^{-2}};\]