Question

A particle of mass m is dropped in a tunnel through the earth (of mass M and radius R). Find its time period of oscillation.

*Assume that the tunnel is not the diameter of the earth (i.e. it is a chord)

*Draw the diagram with proper force vectors in this case.

*Please don't spam.

#Revision Q14​

Answer 1

Answer:

• The time period of the oscillation is T = 2π√(R³/GM)

Explanation:

Let the particle be represented as P, and let the distance b/w the particle and the center of earth be "r". (Refer the diagram)

We know that,

⇒ F = mE

       {E is gravitational intensity of earth at Point say "P".}

And, now as we know that Point p is inside the earth so, gravitational intensity will be E = (GM/R³) . r

⇒ F = m × (GM/R³) . r   ______[1]

Now, from the diagram we can make out that the force is acting at an angle θ then the resultant force will be

⇒ Fᵣ = - F cosθ

⇒ Fᵣ = - F × (x/R)       ∵[cosθ = x/r]

⇒ Fᵣ = - m × (GM/R³) . r × (x/r)

Here, our primary aim is to find acceleration, so

⇒ ma = - m × (GM/R³) . r × (x/r)

⇒ a = - (GM/R³) . r × (x/r)

⇒ a = - (GM/R³) . x

From the formula of Time period.

⇒ T = 2π√(|x/a|)

Substituting the values,

⇒ T = 2π × √(R³/GM)

⇒ T = 2π√(R³/GM)

        (OR)

⇒ T = 2π√(R/g)           ∵{g = GM/R²}

Hence we got the value of time period.

Answer 2

Answer:

The time period of oscillation will be =

$T = 2π√(\frac{R³ }{ GM } )$

Explanation:

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