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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

  • 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 Maximum speed of the particle is

    A
    `sqrt((2GM)/(R ))`
    B
    `sqrt((GM)/(R ))`
    C
    `sqrt((3GM)/(2R ))`
    D
    `sqrt((GM)/(2R ))`

  • If a tunnel is dug along the diameter of the earth and a ball is dropped into the tunnel, it will have

    A
    linear motion
    B
    circular motion
    C
    oscillatory motion
    D
    translatory motion

  • If a tunnel is dug along the diameter of earth and a piece of stone is dropped into it, then the stone will

    A
    come out of the another end of earth and will excape out in space
    B
    come to rest at the centre of earth
    C
    start oscillating about the centre of the earth
    D
    stop at another end of earth