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A horizontal seismometer consists of a pendulum of length "l" constrained to rotate about a nearly vertical axis. The angle away from vertical for the axis is "a" (alpha) and the mass is "m"
a. Determine the Lagrangian for the system in terms of "theta", the angular displacement away from equilibrium about the axis.
b. Find the equation of motion.
c. In the limit of small "theta" find the natural frequency of oscillation
d. If "l" is 30cm, what value of "a" (alpha) will give a period of 10 seconds?

Attached is diagram of Lehman Seismometer.


Solution Preview

alpha (or a) == theta in the attachment,
theta == phi in the attachment
l = L in the attachment

Lagrangian, L = K - V

K = (1/2)*M*l^2 * (dQ/dt)^2 = (1/2)*M*l^2 * w^2 [ w means omega = dQ/dt]

V = M*g*l*sin(a) - M*g*l*sin(a) * cos(Q)
[Assume a tilted rod (by alpha with horizontal) rotated by theta angle. The rod tip ...

Solution Summary

The solution estimates Lagrangian, and also provides the equation of motion and frequency of oscillation and finally angle of inclination for given time period.