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  • Can someone help me with this, If a body is released into a tunnel dug across the diameter of earth, it executes simple harmonic motion with time period

If a body is released into a tunnel dug across the diameter of earth, it executes simple harmonic motion with time period

  • Option 1)

    T = 2\pi \sqrt{\frac{R_{e}}{g}}

  • Option 2)

    T = 2\pi \sqrt{\frac{2R_{e}}{g}}

  • Option 3)

    T = 2\pi \sqrt{\frac{R_{e}}{2g}}

  • Option 4)

    T = 2 \;s

 

Answers (1)

best_answer

Let the bob velocity be v at point B where it makes an angle of 60o with the vertical, then using conservation of mechanical energy 

kE_{A} + PE_{A} = kE_{B} + PE_{B} \\*\Rightarrow \frac{1}{2}m\times 3^{2} = \frac{1}{2}mv^{2} + mgl(1-cos\theta) \\*\Rightarrow 9 = v^{2} + 2\times 10\times 0.5 \frac{1}{2} \Rightarrow v =2m/s

 

Time period of a ball through tunnel in earth -

T= 2\pi \sqrt{\frac{R}{g}}\\=84.6\: min

- wherein

R= Radius of earth

g= acceleration due to gravity

 

 

 


Option 1)

T = 2\pi \sqrt{\frac{R_{e}}{g}}

This is incorrect.

Option 2)

T = 2\pi \sqrt{\frac{2R_{e}}{g}}

This is incorrect.

Option 3)

T = 2\pi \sqrt{\frac{R_{e}}{2g}}

This is incorrect.

Option 4)

T = 2 \;s

This is correct.

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Plabita

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