Q

# Need clarity, kindly explain! - Oscillations and Waves - JEE Main-4

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

• Option 1)

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

• Option 2)

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

• Option 3)

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

• Option 4)

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

110 Views

$T = 2\pi\sqrt{\frac{L}{g_{eff}}} \Rightarrow 2\pi\sqrt{\frac{L}{g\cos \alpha}}$

Time period of simple pendulum accelerating down an incline -

$T=2\pi \sqrt{\frac{l}{g\cos \Theta }}$

- wherein

$l=$ length of pendulum

$g=$ acceleration due to gravity.

$\Theta=$ angle of inclination

Option 1)

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

This is correct.

Option 2)

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

This is incorrect.

Option 3)

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

This is incorrect.

Option 4)

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

This is incorrect.

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