A simple pendulum is being used to determine the value of gravitational acceleration ‘g’ at a certain place. The length of the pendulum is $$20.0$$ cm known to $$1mm$$ accuracy and a stop watch with $$1s$$ resolution measures the time taken for $$40$$ oscillations to be $$50s$$. The accuracy in g is

Question

Infinity Learn · Accepted Answer

$$T=2$$$$π$$lg​on squaring both sides​$$g=4$$$$π$$$$2lT2$$​on differentiating,​Δ$$gg=$$Δ$$ll+2$$Δ$$TT=0.120+2$$×$$150=0.045$$ ⇒Δgg×$$100=0.045$$​ ×$$100$$        ∴%Δ$$gg=4.5$$%

A simple pendulum is being used to determine the value of gravitational acceleration ‘g’ at a certain place. The length of the pendulum is 20.0 cm known to 1mm accuracy and a stop watch with 1s resolution measures the time taken for 40 oscillations to be 50s. The accuracy in g is

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