Question

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 $\alpha$ is given by

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

The acceleration of the vehicle down the plane is g sin $\alpha$ The reaction force acting on the penduclum bob gives it an acceleration a = g sin $\alpha$ up the plane. This acceleration has two rectangular components $a_x=acos\alpha =gsin\alpha cos\alpha and{a}_y=asin\alpha =gsi{n}^2\alpha$ as shown in fig. 9.36. the effective acceleration due to gravity acting on the bob is given by

$g_{eff}^2=a_x^2+(g-a_y{)}^2$
$=a_x^2+g^2+a_y^2-2g{a}_y$
$=g^{2}si{n}^2\alpha co{s}^2\alpha +g^2+g^{2}so{m}^4\alpha -2g^{2}si{n}^2\alpha$
$=g^{2}si{n}^2\alpha (co{s}^2\alpha +si{n}^2\alpha )+g^{2}si{n}^2\alpha$
$=g^2(1-si{n}^2\alpha )=g^{2}co{s}^2\alpha$
$Or{g}_{eff}=gcos\alpha$
$NowT=2\pi \sqrt{\frac{L}{g_{eff}}}=2\pi \sqrt{\frac{L}{gcos\alpha }}$
Hence the correct choice is (a)