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 :
$$2\pi \sqrt {\frac{L}{{g\cos \alpha }}} $$
Solution :
As shown in the figure, $$g\sin \alpha $$ is the pseudo acceleration applied by the observer in the accelerated frame
$$\eqalign{ & {a_y} = g - g{\sin ^2}\alpha = g\left( {1 - {{\sin }^2}\alpha } \right) = g{\cos ^2}\alpha \cr & a = \sqrt {a_x^2 + a_y^2} \cr & = \sqrt {{g^2}{{\sin }^2}\alpha {{\cos }^2}\alpha + {g^2}{{\cos }^4}\alpha } \cr & = g\cos \alpha \sqrt {{{\sin }^2}\alpha + {{\cos }^2}\alpha } = g\cos \alpha \cr & \therefore T = 2\pi \sqrt {\frac{L}{{g\cos \alpha }}} \cr} $$
NOTE: Whenever point of suspension is accelerating use $$T = 2\pi \sqrt {\frac{L}{{{g_{eff{\text{ }}}}}}} $$
As shown in the figure, $$g\sin \alpha $$ is the pseudo acceleration applied by the observer in the accelerated frame
$$\eqalign{ & {a_y} = g - g{\sin ^2}\alpha = g\left( {1 - {{\sin }^2}\alpha } \right) = g{\cos ^2}\alpha \cr & a = \sqrt {a_x^2 + a_y^2} \cr & = \sqrt {{g^2}{{\sin }^2}\alpha {{\cos }^2}\alpha + {g^2}{{\cos }^4}\alpha } \cr & = g\cos \alpha \sqrt {{{\sin }^2}\alpha + {{\cos }^2}\alpha } = g\cos \alpha \cr & \therefore T = 2\pi \sqrt {\frac{L}{{g\cos \alpha }}} \cr} $$
NOTE: Whenever point of suspension is accelerating use $$T = 2\pi \sqrt {\frac{L}{{{g_{eff{\text{ }}}}}}} $$