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. $2\pi \sqrt{\frac{L}{g}}$
• C. $2\pi \sqrt{\frac{L}{g\sin \alpha }}$
• D. $2\pi \sqrt{\frac{L}{g\tan \alpha }}$

Answer 1

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

$g_{eff}^2=a_x^2+(g-a_y{)}^2$
$⟹g_{eff}^2=a_x^2+g^2+a_y^2-2g{a}_y$
$⟹g_{eff}^2=g^2{\sin }^2\alpha {\cos }^2\alpha +g^2+g^2{\sin }^4\alpha -2g^2{\sin }^2\alpha$
$⟹g_{eff}^2=g^2(1-si{n}^2\alpha )=g^2{\cos }^2\alpha$
$⟹g_{eff}=g\cos \alpha$
Now, $T=2\pi \sqrt{\frac{L}{g_{eff}}}=2\pi \sqrt{\frac{L}{g\cos \alpha }}$