Question

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

• A. 3%
• B. 1%
• C. 5%
• D. 2%

Answer 1

The correct option is A. 3%.
Given,

$\Delta L=1mm=0.001m$

$L=20cm$

$\Delta T=\frac{Leastcount}{No.ofoscillations}=\frac{1}{100}=0.01s$

$T=\frac{Totaltime}{No.ofoscillations}=\frac{90}{100}=0.9s$

From the given equation $g$ can be written as,
$g=\frac{4{\pi }^{2}L}{T^2}$

The relative error in $g$ is given by,
$\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\left(\frac{\Delta T}{T}\right)$

$\frac{\Delta L}{L}=\frac{0.001}{0.20},\frac{\Delta T}{T}=\frac{0.01}{0.9}$

$100\left(\frac{\Delta g}{g}\right)=[\left(\frac{0.001}{0.20}\right)+2\times \left(\frac{0.01}{0.9}\right)]\times 100\%=2.7\%|\simeq 3\%$.