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If a smooth tunnel is dug across a diame...

If a smooth tunnel is dug across a diameter of earth and a particle is released from the surface of earth, the particle oscillate simple harmonically along it
Time period of the particle is not equal to

Text Solution

Verified by Experts

Force acting on the particle `=(m_o)(g_"earth")`
`=(m_o)((Gm)/R^3)x`
As this form is opposite of x so we can write
`F=-((Gm m_o)/R^3)x`
Now this form `F alpha-x`, So motion of the particle Will be simple harmonia motion
`F=-((Gm m_o)/R^3)x`
F=-Kx
Comparing with the standard eqn. of SHM the force constant `K=(Gm m_o)/R^3`
So time period of the particle .
`T=2pisqrt(m_o/k)`
`T=2pisqrt((m_o/(Gm_(mo)))/R^3)`
`T=2pisqrt(R^3/(Gm))`
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Knowledge Check

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