A 1,200 kg earthquake detector responds to a force of 1.2 x 10 ^ (6) Newtons at

Question

A 1,200 kg earthquake detector responds to a force of 1.2 x 10 ^ (6) Newtons at its resonant frequency of .092 radians/sec with an amplitude of 1.1mm. What is the bandwidth (Full Width Half Maximum) of its response?

My Professors Work and Answer:

(1.2 x 10^6) x (sqrt (3)) / ((.0011) x (.092) x (1,200)) = 1.71 x 10 ^5

The formula my professor gave me for FWHM (Full Width Half Maximum):

FWHM / (resonant frequency) = (Driving Force) x (sqrt(3)) / (m * resonant frequency )

He also wrote up this formula on the board which contradicts the previous:

FWHM / (resonant frequency) = (Driving Force) x (sqrt(3)) / (m * Driving Force frequency )

Based off the work above it looks like the following is going on and I'm quite confused:

FWHM = ((Driving Force) x (sqrt 3)) / ((Amplitude) x (resonant frequency) x m)

Explanation / Answer

In case of such confusion go back to basics of dimenssional notation

For bandwidth SI unit is per second i.e. T-1

For force dimentional notation is MLT-2 where M = mass L = length

so we need to divide MLT-2 such that we get T-1

MLT-2 / (M*L*T-1) which gives us force/(mass*amplitude*freq)

so clearly formulas you sir gave are not correct dimensionally

I suggest you to point this out to him.

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