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प्रश्न
A tunnel is dug through the centre of the Earth. Show that a body of mass ‘m’ when dropped from rest from one end of the tunnel will execute simple harmonic motion.
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उत्तर
Consider the situation shown in the diagram.
The gravitational force on the particle at a distance r from the centre of the earth arises entirely from that portion of matter of the earth in shells internal to the position of the particle. The external shells exert no force on the particle.
More clearly,
Let g' be the acceleration at P.
So, `g^' = g(1 - d/R) = g((R - d)/R)`
From the figure, `R - d = y`
⇒ `g^' = g y/R^'`
Force on the body at p,
F = `- mg^' = (- mg)/R y` .......(i)
⇒ F ∝ – y ......[Where y is the distance from the centre]
So, motion is S.H.M.
For time period, we can write equation (i)
As ma = `- (Mg)/Ry`
⇒ `a = - g/R y`
Comparing with a = `- ω^2y`
`ω^2 = g/R`
⇒ `((2pi)/T) = g/R`
⇒ T = `2pi sqrt(R/g)`
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संबंधित प्रश्न
let us take the position of mass when the spring is unstretched as x = 0, and the direction from left to right as the positive direction of the x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is
(a) at the mean position,
(b) at the maximum stretched position, and
(c) at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?
Answer the following questions:
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