The period of oscillation of a simple pendulum is $$T=2$$$$π$$Lg Measured value of L is $$20$$ cm known to $$1mm$$ accuracy and time for $$100$$ oscillations of the pendulum is found to be $$90$$ s using a $$wrist-watch$$ of $$1$$ s resolution. The accuracy in the determination of g is

Question

The period of oscillation of a simple pendulum is $$T=2$$$$π$$Lg Measured value of L is $$20$$ cm known to $$1mm$$ accuracy and time for $$100$$ oscillations of the pendulum is found to be $$90$$ s using a $$wrist-watch$$ of $$1$$ s resolution. The accuracy in the determination of g is

Infinity Learn · Accepted Answer

Given, $$T=2$$$$π$$Lgor $$g=4$$$$π$$$$2LT2Hence$$ $$T=tn$$ and Δ$$T=$$Δtn..            Δ$$TT=$$ΔttThe errors in both L and t are the least count errors. Therefore,Δ$$gg=$$Δ$$LL+2$$Δ$$TT=0.120.0+2190=0.027The$$ percentage error in g isΔgg×$$100=$$ΔLL×$$100+2$$×ΔTT×$$100=3$$%

The period of oscillation of a simple pendulum is T $= 2 π \sqrt{\frac{L}{g}}$ Measured value of L is 20 cm known to 1mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist-watch of 1 s resolution. The accuracy in the determination of g is

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The period of oscillation of a simple pendulum is T=2πLg Measured value of L is 20 cm known to 1mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist-watch of 1 s resolution. The accuracy in the determination of g is

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Ready to Test Your Skills?

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The period of oscillation of a simple pendulum is T $= 2 π \sqrt{\frac{L}{g}}$ Measured value of L is 20 cm known to 1mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist-watch of 1 s resolution. The accuracy in the determination of g is