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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}}

 
Answers (1)
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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