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प्रश्न
Consider the Earth as a homogeneous sphere of radius R and a straight hole is bored in it through its centre. Show that a particle dropped into the hole will execute a simple harmonic motion such that its time period is
T = `2π sqrt("R"/"g")`
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उत्तर
Oscillations of a particle dropped in a tunnel along the diameter of the earth.
Consider earth to be a sphere of radius R and centre O. A straight tunnel is dug along the diameter of the earth. Let ‘g’ be the value of acceleration due to gravity at the surface of the earth.
Suppose a body of mass ‘m’ is dropped into the tunnel and it is at point R i.e., at a depth d below the surface of the earth at any instant.
If g’ is the acceleration due to gravity at P.
then R − d = y
∴ g’ = `"g"("y"/"R")`
Force acting on the body a point P is
F = − mg’ = `-"mg"/"R""y"` i.e., `"F" ∝ "y"`
Negative sign indicates that the force acts in the opposite direction of displacement.
Thus the body will execute SHM with force constant, k = `"mg"/"R"`
The period of oscillation of the body will be T = `2π sqrt("m"/"k") = 2π sqrt ("m"/("mg"//"R"))`
T = `2π sqrt("R"/"g")`
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