The period of oscillation of a simple pendulum is $$T=2$$$$π$$$$L/g$$. Measured value of L is $$20.0$$ cm known to $$1$$ mm accuracy and time for $$100$$ oscillations of the pendulum is found to be $$90$$ s using a wrist watch of $$1s$$ resolution. What is the accuracy in the determination of g?

Question

The period of oscillation of a simple pendulum is $$T=2$$$$π$$$$L/g$$. Measured value of L is $$20.0$$ cm known to $$1$$ mm accuracy and time for $$100$$ oscillations of the pendulum is found to be $$90$$ s using a wrist watch of $$1s$$ resolution. What is the accuracy in the determination of g?

Infinity Learn · Accepted Answer

$$g=4$$$$π$$$$2L/T2Here$$, $$T=tn$$ and Δ$$T=$$ΔtnTherefore, Δ$$TT=$$ΔttThe error in both L and t are least count error,Therefore,$$($$Δ$$g/g)=($$Δ$$L/L)+2($$Δ$$T/T)=($$Δ$$L/L)+2($$Δ$$t/t)=0.120.0+2190=0.027Thus$$, the percentage error in g is,$$($$Δ$$g/g)$$×$$100=2.7$$%

The period of oscillation of a simple pendulum is $T = 2 π \sqrt{L / g}$ . Measured value of L is $20.0 cm$ known to $1 mm$ accuracy and time for 100 oscillations of the pendulum is found to be $90 s$ using a wrist watch of 1s resolution. What is the accuracy in the determination of g?

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The period of oscillation of a simple pendulum is T=2πL/g. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1s resolution. What is the accuracy in the determination of g?

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

The period of oscillation of a simple pendulum is $T = 2 π \sqrt{L / g}$ . Measured value of L is $20.0 cm$ known to $1 mm$ accuracy and time for 100 oscillations of the pendulum is found to be $90 s$ using a wrist watch of 1s resolution. What is the accuracy in the determination of g?