Question

The time 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 α, is given by:

• A. $2\pi\sqrt {\frac {L}{(g\ cos⁡\ α)}}$
• B. $2\pi\sqrt {\frac {L}{(g\ sin\ ⁡α)}}$
• C. $2\pi\sqrt {\frac Lg}$
• D. $2π\sqrt {\frac {L}{(g\ tan\ ⁡α)}}$

Answer 1

Answer: (A)

$2\pi\sqrt {\frac {L}{(g\ cos⁡\ α)}}$

Answer 2

$|g_{eff}|=|\bar g−\bar a|$
$⇒ g_{eff} = g\ cos\ α$
The time period,
$⇒ T=2\pi \sqrt {\frac {I}{g_{eff}}}$

$T=2\pi\sqrt {\frac {L}{g\ cos⁡\ α}}$

So, the correct option is (A): $2\pi\sqrt {\frac {L}{(g\ cos⁡\ α)}}$