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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Subjective Medium Published on 18th 08, 2020
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