Question

The period of oscillation of simple pendulum of length $l$ suspended from the roof of the vehicle which moves without friction, down an inclined plane of inclination $\alpha$, is given by

• A. $2\pi\sqrt{\frac{L}{g cos \alpha}}$
• 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 1

Answer: (A)

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

Answer 2

We are given that the simple pendulum of length $l$ is hanging from the roof of a vehicle which is moving down the frictionless inclined plane.
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 we can say that the effective acceleration would be
equal to
$g_{eff} = g\, cos \,\alpha$
Now the time period of oscillation is given by
$T =2\pi \sqrt{\frac{l}{g_{eff}}}=2\pi \sqrt{\frac{l}{g \,cos\, a}}$