The period of oscillation of a simple pendulum of length ‘L’ suspended from the ceiling of a vehicle which moves without friction down a fixed inclined plane of inclination$\text{'}\alpha \text{'}$  is given by

Due to the frame getting an acceleration$\left(\text{g}\sin \theta \right)$  downwards, there will be a pseudo force $\left(\text{mg}\sin \alpha \right)$ on the bob, at an angle$\left(\alpha +\frac{\pi }{2}\right)$  from vertical downward ‘g’.

The magnitude of resultant acceleration${\text{a}}_{\text{1}}=\text{g}$  

${\text{a}}_{\text{2}}=\text{gsin}\alpha \left(\because {\alpha }^{|}=\alpha +\frac{\pi }{2}\right)$

$\text{a}=\sqrt{{\text{a}}_{\text{1}}^{\text{2}}+{\text{a}}_{\text{2}}^{\text{2}}+{\text{a}}_{\text{1}}{\text{a}}_{\text{2}}\cos {\theta }^{|}}$

$\text{a}=\sqrt{{\text{g}}^{\text{2}}+{\text{g}}^{\text{2}}{\sin }^2\alpha +2{\text{g}}^{\text{2}}\sin \alpha \cos \left(\frac{\pi }{2}+\alpha \right)}$

$\text{a}=\sqrt{{\text{g}}^{\text{2}}+{\text{g}}^{\text{2}}{\sin }^2\alpha -2{\text{g}}^{\text{2}}{\sin }^2\alpha }\Rightarrow \text{a}=\text{g}\cos \alpha$

$\text{a}=\sqrt{{\text{g}}^{\text{2}}-{\text{g}}^{\text{2}}{\sin }^2\alpha }=\sqrt{{\text{g}}^{\text{2}}\left(1-{\sin }^2\alpha \right)}$

$\text{T}=2\pi \sqrt{\frac{\text{L}}{\text{a}}}=2\pi \sqrt{\frac{\text{L}}{\text{g}\cos \alpha }}$

$\text{a}=\sqrt{{\text{g}}^{\text{2}}{\cos }^2\alpha }=\text{g}\cos \alpha$