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Question

Assume that a tunnel is dug across the earth (radius = R) passing through its centre. Find the time a particle takes to cover the length of the tunnel if (a) it is projected into the tunnel with a speed of gR (b) it is released from a height R above the tunnel (c) it is thrown vertically upward along the length of tunnel with a speed of gR

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Solution

Let M be the total mass of the earth.

At any position x.

M1M=p×43πx3p×43πr3×x3R3

M1=Mx3R3

So, force on the particle is given by

Fx=GMmx2

=GMmR3x ...(1)

So, acceleration of the mass 'M' at that position is given by,

ax=GMR3

axx=ω2

=GMR3=gR (g=GMR2)

So, T=2π (Rg) = Time period of oscillation

(a) Now using velocity = displacement equation

v=ω (A2Y2) [where, A = amplitude]

Given when,

y=R, v=gR,w=gR

gR=(gR)(A2R2)

[because ω=kg/r]

R2=A2R2

A=2R

[Now, the phase of the particle at the point P is greater than π2 has less than π and at Q is greater than π but less than 3π2. Let the times taken by the particle to reach the positions P and Q be t1 and t2 respectively, then using displacement equation].

y=r sin ωt

We have,

R=2 R sin ωt

ωt1=3π4

We have,

R=2 R sin ωt1

ωt2=5π4

So, ω (t2t1)=π2

t2t1=π2ωπ2(Rg)

Time taken by the particle to travel from P to Q is

t2t1=π2(Rg) sec.

(b) When the body is dropped from a height r, then applying conservation of energy, change in P.E. = gain in K.E.

GMmRGMm2R=12mv2

v=(gR)

Since, the velocity is same at P, as in part (a) the body will take same time to travel PQ.

(c) When the body is projected vertically upward from P with a velocity.

(gR) its velocity will be zero at the highest point.

The velocity of the body, when reaches p, again will be

v=(gR)

Hence the body will take same time

π2 (Rg) to travel P.Q.


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