The period of oscillation of a simple pendulum of length L suspen… - Vidyadip

MCQ

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

✓
$2\pi \sqrt {\frac{L}{{g\cos \alpha }}} $
B
$2\pi \sqrt {\frac{L}{{g\sin \alpha }}} $
C
$2\pi \sqrt {\frac{L}{g}} $
D
$2\pi \sqrt {\frac{L}{{g\tan \alpha }}} $

✓

Answer

Correct option: A.

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

a
(a) See the following force diagram. Vehicle is moving down the frictionless inclined surface so, it's acceleration is $g\sin \theta $. Since vehicle is accelerating, a pseudo force $m(g\sin \theta )$ will act on bob of pendulum which cancel the $\sin \theta $ component of weight of the bob.

Hence net force on the bob is $F_{net} = mg\cos \theta $ or net acceleration of the bob is ${g_{eff}} = g\cos \theta $

$\therefore $ Time period $T = 2\pi \sqrt {\frac{l}{{{g_{eff}}}}} = 2\pi \sqrt {\frac{l}{{g\cos \theta }}} $

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