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 $60^{\circ}$ is given by $π \sqrt{\frac{X L}{g}}$ then find $X$ .

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Solution

As we know,
$T = 2 π \sqrt{\frac{L}{g_{e f f}}} T = 2 π \sqrt{\frac{L}{g c o s 60}} T = 2 π \sqrt{\frac{2 L}{g}}$

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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 60∘ is given by π√XLg then find X.

As we know, T=2π√LgeffT=2π√Lgcos60T=2π√2Lg

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 $60^{\circ}$ is given by $π \sqrt{\frac{X L}{g}}$ then find $X$ .

As we know,
$T = 2 π \sqrt{\frac{L}{g_{e f f}}} T = 2 π \sqrt{\frac{L}{g c o s 60}} T = 2 π \sqrt{\frac{2 L}{g}}$