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Mechanical Properties Of Fluids

Question
CBSEENPH11019242

Show that motion of body dropped in a tunnel dug along the diameter of the earth is simple harmonic motion. Find the time period of motion.

Solution
Let the earth be a homogenous sphere of radius R and density ρ.
Let the bore be dug along the diameter of the earth.
                 
The acceleration due to gravity at point P at a distance y from the centre of the earth is,
straight a equals 4 over 3 πGpy equals 4 over 3 πGρ straight R over straight R straight y equals straight g over straight R straight y
w here comma space straight g equals 4 over 3 πGρR, is the acceleration due to gravity at the surface of the earth.
Since acceleration 'a' at point P is directed towards the centre of the earth, therefore in vector form we can write, 
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Therefore the motion is simple harmonic motion.

space space space space space space straight omega squared equals straight g over straight R
therefore space space space straight T equals 2 straight pi square root of 1 over straight omega squared end root
space space space space space space space space space equals 2 straight pi square root of straight R over straight g end root 
Substituting R and g, we get,
T equals 2 pi square root of fraction numerator 6.4 cross times 10 to the power of 6 over denominator 9.8 end fraction end root equals 5079 s, is the time period of the motion.