A simple seismograph consists of a mass M hung from a spring on a rigid framewor

Question

A simple seismograph consists of a mass M hung from a spring on a rigid framework attached to the Earth. The spring constant is k, and there is a damping force that depends on the velocity of the mass relative to the Earth. Supposing that there is a vertical displacement h of the ground at time t. Show that the extension of the spring y is given by: d^2y/dt^2 + 2 gamma dy/dt + omega^2 y = d^2 h/dt^2 Determine the response of the seismograph to a harmonic seismic wave described by h = C cos ohm t. A typical seismograph has a period r = 30s, and a Q factor of As a result of an earthquake a seismic wave is produced with a period of 30 minutes, and an amplitude such that the maximum vertical acceleration of the ground is 10^-9m/s^2. What is the amplitude of the corresponding oscillations of the mass in the seismograph?

Explanation / Answer

Solution :-

a)   Let r0 = eq. distance of the earth’s surface with respect to the fixed star. And y0 = eq. distance of M with respect to earth’s surface.

h = displacement of Earth’s surface with respect to fixed star.

y = displacement of M (mass) with respect to the earth’s surface.

Hence the displacement of M with respect to relative stars is shown below:-

= (r0 + h + y0 + y) (r0 + y0) = h + y.

Now we have to find acceleration with respect to fixed stars,

Acceleration, a = d2/dt2 () = d2/dt2 (h ) + d2/dt2 (y)

So now we will write the force equation

That is M (d2/dt2 (h) + d2/dt2 (y)) = F (damping) + F(spring)

Here, F (damping) = -r dy/dt

And F (spring) = -2ky

Note: - r = Proportionality constant and k = Spring Constant.

Let us now define this 0^2 = k/M

And = r/M.

Hence the equation becomes: - d2y/dt2 + 2 dy/dt + 0^2y = - d2h/dt.

b) So if h = C cos t then equation q becomes,

d2y/dt2 + dy/dt + 0^2y = C ^2e^(i t)

y˜ = A˜ e^(i t)

So A˜ = C ^2/ (0^2- ^2+ i )

= (C ^2/ Sqrt ((0^2- ^2)^2+ 2 2)) ei

Here

= tan ^ (-1) ( / 0^2- ^2)

Hence the required steady state vibration equation will become

y = Re {y}

= [C ^2/ Sqrt [ (^2- ^2) + ^2^2)]] Cos (t – )

c) Quality Factor = 0/2* = 2*pi() / To for 0 = 2*pi() / To and = /2

Given Qo = 2 and

To = 30.0 sec

So = pi()/ 30 sec^-1

So the time period of the oscillation of the Earth’s surface is:-

T = 2*pi()/ = 20 minutes and its maximum acceleration C ^2 = 10^-9 m/sec^2

So using the formula,

A(),   The amplitude of the corresponding oscillation of the mass in the seismograph, A = 0.023 m.

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