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

• Correct answer
  A

  $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}\cos \mathrm{\alpha }}}$
• B

  $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}\sin \mathrm{\alpha }}}$
• C

  $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$
• D

  $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}\tan \mathrm{\alpha }}}$

Solution

When vehicle is accelerated ${\mathrm{F}}_{\mathrm{pseudo}}$ will act on pendulum which will balance the force along the incline.

${\mathrm{F}}_{\mathrm{P}}=\mathrm{m}\mathrm{g}\sin \mathrm{\alpha }$

Therefore, ${\mathrm{g}}_{\mathrm{eff}}=\mathrm{gcos}\mathrm{\alpha }$.

$\therefore \mathrm{ }\mathrm{ }\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}\cos \mathrm{\alpha }}}$

Hence, the correct answer is option: 'A'.