Question

8, Consider a pendulum bob of length 10cm hung from a ceiling. What should we do to the length in order to vary the period by 2?

Answer 1

We know that, $T=2\pi \sqrt{\frac{L}{g}}$

Here, $L=10 \ cm$ $, g=9.8 m/s^2$

So, $T=2\pi\sqrt{\frac{10}{9.8}}=6.347\ s$

If $T$ is varying by 2 then new length $(L_1)$ will be calculated by

$T\pm2=2\pi \sqrt{\frac{L_1}{g}}$

$\Rightarrow 6.347\pm2=2\pi \sqrt{\frac{L_1}{9.8}}$

For $+$ ve sign

$6.347+2=2\pi \sqrt{\frac{L_1}{9.8}} \\\Rightarrow L_1=17.29\ m$

For $-$ ve sign

$6.347-2=2\pi \sqrt{\frac{L_1}{9.8}} \\\Rightarrow L_1=4.69\ m$