Discussion Forum

The period of oscillation of a simple pendulam 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 pendulam is found to be 90 s using  a wrist watch of 1 sresolution .The accurancy in the determination of g is

Answer:

Explanation:

Central idea :Given time period $T= 2\pi \sqrt{\frac{L}{g}}$ 

Thus,Changes can be expressed as 

$\pm \frac{2\Delta T}{T}=\pm \frac{\Delta L}{L}\pm \frac{\Delta g}{g}$

According to the question, we can write

$\frac{\Delta L}{L}=\frac{0.1 cm}{20.0 cm}= \frac{1}{200}$

Again time period

$T= \frac{90}{100} s $and $\Delta T=\frac{1}{100} s$

$\frac{\Delta  T}{T}=\frac{1}{90} s$

Now,

$T=2\pi \sqrt{\frac{L}{g}}$

$g=4\pi ^{2}\frac{L}{T^{2}}$

$\frac{\Delta g}{g}=\frac{\Delta L}{L} + \frac{2\Delta T}{T}$

$\frac{\Delta g}{g}\times 100%=\frac{\Delta L}{L}\times 100% + \frac{2\Delta T}{T}\times 100%$

$=\left ( \frac{1}{200}\times 100 \right )% + 2\times \frac{1}{90}\times 100% $

$\simeq2.72 \simeq 3%$