Physics

The period of oscillation of a simple pendulum is $$T = 2\pi \sqrt {L/g} .$$ 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 1 sec resolution. What is the accuracy in the determination of  g?


SOLUTION
For the determination for the value of G 
we require the formula for time period 
$$\therefore T=2\pi \sqrt{\dfrac{L}{g}}$$
$$L=20cm =0.2m$$
$$\Delta L=1mm=0.01m$$
Time for 100 oscilation is 90s.
so, $$T=0.9s$$
Now, 
$$T=2\pi \sqrt{\dfrac{L}{g}}$$
$$\Rightarrow g=4\pi^2\times \dfrac{L}{g}$$
$$\Rightarrow \dfrac{\Delta g}{g}=\dfrac{\Delta L}{L}+\dfrac{\Delta T}{T}$$
$$\Rightarrow \dfrac{\Delta g }{g}=\dfrac{0.001}{0.2}+\dfrac{0.01}{0.9}$$
$$\Rightarrow \dfrac{\Delta g }{g}=0.016$$
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