Question

The period of oscillation of a simple pendulum is t=2π√(l/g). Measured value of l=20cm known to 1mm accuracy and the for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1s resolution. What is the accuracy in the determination of g?

Answer 1

Answer:

hope this answer of mine helps you

Answer 2

Answer: The accuracy in the determination of g is ±2.7%

Explanation: Given, $t=2\pi \sqrt{\frac{l}{g}}$ ...(1)

l= 20 cm known to have 1 mm accuracy

$\therefore$$\Delta l = 1mm$

t= 90s known to a resolution of 1s

$\therefore\Delta t=1s$

Taking Eqn (1) and squaring on both sides,

$t^{2} = 4\pi ^{2} \frac{l}{g}$

Cross multiply,

$g=4\pi^{2} \frac{l}{t^{2} }$

Take ln on both sides and differentiate,$\frac{\Delta g}{g}=\frac{\Delta l}{l} + 2\frac{\Delta t}{t}$

Putting all the values,$\frac{\Delta g}{g}=\frac{0.1}{20}+2*\frac{1}{90}$ = 2.7

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