Question

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

• 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

The correct option is A $2\pi \sqrt{\frac{L}{g\cos \alpha }}$

Since the vehicle is moving down the plane with an acceleration
$g\sin \alpha$ and the bob moves with same acceleration (because, each and every particle of the system is accelerating with same acceleration down the plane).
$\Rightarrow$ When the bob is released, it moves with effective acceleration of
$g_{eff}=\sqrt{g^2-g^2{\sin }^2\alpha }=g\cos \alpha$
$\Rightarrow$ The period of oscillation of the pendulum
$=2\pi \sqrt{\frac{l}{g^{{}^{\prime }}}}=2\pi \sqrt{\frac{l}{g\cos \alpha }}$

OR

As pendulum is in non-inertial frame,
Applying psuedo force $F_p=mg\sin \alpha$, along the plane,
On diving $mg$ into components, we get $mg\sin \alpha$ along the plane and $mg\cos \alpha$ perpendicular to plane.
Hence along the plane $F_p$ and $mg\sin \alpha$ balance each other and $mg\cos \alpha$ causes oscillation
Effective acceleration is $a=g\cos \alpha$
$\Rightarrow$ The period of oscillation of the pendulum
$=2\pi \sqrt{\frac{l}{g^{{}^{\prime }}}}=2\pi \sqrt{\frac{l}{g\cos \alpha }}$