Question

The period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle which moves without friction down in an inclined plane of inclination $\alpha ,$ is given by

• A. $2\pi \sqrt{\frac{L}{gcos\alpha }}$
• B. $2\pi \sqrt{\frac{L}{gsin\alpha }}$
• C. $2\pi \sqrt{\frac{L}{g}}$
• D. $2\pi \sqrt{\frac{L}{gtan\alpha }}$

Answer 1

The correct option is A $2\pi \sqrt{\frac{L}{gcos\alpha }}$

$a_n=gsin\alpha cos\alpha$
$a_y=g-gsi{n}^2\alpha =g(1-si{n}^2\alpha )=gco{s}^2\alpha$
$a=\sqrt{a_x^2+a_y^2}=\sqrt{g^{2}si{n}^2\alpha +g^{2}co{s}^4\alpha }$
$=gcos\sqrt{si{n}^2\alpha +co{s}^2\alpha }=gcos\alpha$
$\therefore T=2\pi \sqrt{\frac{L}{gcos\alpha }}$