Explanation

(a) The acceleration of the vehicle down the plane$= g\sin\alpha$.
The reaction force acting on the bob of pendulum gives it an acceleration $a( = g\sin\alpha)$ up the plane. This acceleration has two rectangular components,
$a_{x} = a\cos{\alpha = g\sin\alpha\cos\alpha}$
And $a_{y} = a\sin\alpha = g\ \sin^{2}\alpha$ as shown in figure. The effective acceleration due to gravity acting on the bob is given by
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$g_{eff}^{2} = a_{x}^{2} + \left( g - a_{y} \right)^{2} = a_{x}^{2}g^{2} + a_{y}^{2} - 2ga_{y}$
$= g^{2}\sin^{2}\alpha\cos^{2}\alpha + g^{2}\sin^{4}\alpha - 2g \times g\sin^{2}\alpha$
$= g^{2}\sin^{2}\alpha\left( \cos^{2}\alpha + \sin^{2}\alpha \right) + g^{2} - 2g^{2}\sin^{2}\alpha$
$= g^{2}\sin^{2}\alpha + g^{2} - 2g^{2}\sin^{2}\alpha = g^{2}(1 - \sin^{2}\alpha)$
$= g^{2}\cos^{2}\alpha$
$\therefore\ \ \ \ g_{eff} = g\cos\alpha$
Now, $T^{'} = 2\pi\sqrt{\frac{L}{e_{eff}}}$
$= 2\pi\sqrt{\frac{L}{g\cos\alpha}}$