Question

A simple pendulum of length 40 cm is taken inside a deep mine. Assume for the time being that the mine is 1600 km deep, calculate the time period of the pendulum there. Radius of the earth = 6400 km.

Answer 1

Given, Radius of the earth = 6400 km

h = 1600 km,

l = 0.40 m  

''Let M' is the mass of the earth part with radius R-h. ''

At an depth of h, the Force on the bob of the pendulum if its mass is m   = GM'm/(R-h)²

Let the acceleration due to the gravity at an depth of h be g'

then, mg' =GM'm/(R-h)²

Thus, g' = GM'/(R-h)²

Also,  M' = 4π(R-h)³d/3   [Where, d is the density of earth.]

Since, density is always constant, thus,

mass of the earth = M = 4πR³d/3    [Note, Earth is assumed to be an sphere.]

Thus,   M/M' = R³/(R-h)³     ---eq(i)

Since, At the surface of the earth , g = GM/R²

∴ g/g' = (GM/R²)/{GM'/(R-h)²}

∴ g/g' = M(R-h)²/M'R²

∴ g/g' = M/M' × (R - h)²/R²

∴ g/g' = R³/(R - h)³ × (R - h)²/R²

∴ g/g' = R/(R - h)

Now, Since, T  = k/√g   [k = 2π√l]

T' =  k/√g'

∴ T/T' = g'/g

      OR

T'/T = √(g/g')

∴ T'/T = √(R/(R - h))

∴ T' = T × √(6400/(6400 - 1600))

 ∴ T' = T × √(6400/4800)

∴ T' = T × √(8/6)

∴ T' = T × √(4/3)

Now, $T = 2\pi \sqrt{\frac{l}{g} }$

∴ T = 2π × √(0.4/9.8)

∴ T = 1.269 seconds.

∴ T' = T × √(4/3)

∴ T' = 1.47 seconds.

Hence, the time period of the pendulum at an depth of 1600 km is 1.47 seconds.

Hope it helps.

Answer 2

Answer:

Radius of the earth = 6400 km

h = 1600 km,

l = 0.40 m  

''Let M' is the mass of the earth part with radius R-h. ''

At an depth of h, the Force on the bob of the pendulum if its mass is m   = GM'm/(R-h)²

Let the acceleration due to the gravity at an depth of h be g'

then, mg' =GM'm/(R-h)²

Thus, g' = GM'/(R-h)²

Also,  M' = 4π(R-h)³d/3   [Where, d is the density of earth.]

Since, density is always constant, thus,

mass of the earth = M = 4πR³d/3    [Note, Earth is assumed to be an sphere.]

Thus,   M/M' = R³/(R-h)³     ---eq(i)

Since, At the surface of the earth , g = GM/R²

∴ g/g' = (GM/R²)/{GM'/(R-h)²}

∴ g/g' = M(R-h)²/M'R²

∴ g/g' = M/M' × (R - h)²/R²

∴ g/g' = R³/(R - h)³ × (R - h)²/R²

∴ g/g' = R/(R - h)

Now, Since, T  = k/√g   [k = 2π√l]

T' =  k/√g'

∴ T/T' = g'/g

      OR

T'/T = √(g/g')

∴ T'/T = √(R/(R - h))

∴ T' = T × √(6400/(6400 - 1600))

 ∴ T' = T × √(6400/4800)

∴ T' = T × √(8/6)

∴ T' = T × √(4/3)

Now,

∴ T = 2π × √(0.4/9.8)

∴ T = 1.269 seconds.

∴ T' = T × √(4/3)

∴ T' = 1.47 seconds.