Question

Find the period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination Alpha.

Answer 1

The acceleration of the vehicle down the plane is g sin $\alpha$. The reaction force acting on the pendulum bob gives it an accelertion a =g sin $\alpha$ up the plane. This acceleration has two rectangular components $a_x=acos\alpha =gsin\alpha cos\alpha$ and ${\alpha }_y=asin\alpha =gsi{n}^2\alpha$ as shown in Fig. 13.6
The effective acceleration due to gravity acting on the bob is given by
$g_{eff}^2=a_x^2+(g-a_y{)}^2=a_x^2+g^2+a_y^2-2g{a}_y=g^{2}si{n}^2\alpha co{s}^2\alpha +g^2+g^{2}si{n}^4-2g^{2}si{n}^2\alpha$

$=g^{2}si{n}^2\alpha (co{s}^2\alpha +si{n}^2\alpha )+g^2-2g^{2}si{n}^2\alpha =g^2(1-si{n}^2\alpha )=g^{2}co{s}^2\alpha or{g}_{eff}=gcos\alpha NowT=2\pi \sqrt{\frac{L}{g_{eff}}}=2\pi \sqrt{\frac{L}{gcos\alpha }}$