It has recently become possible to \"weigh\" DNA molecules by measuring the infl

Question

It has recently become possible to "weigh" DNA molecules by measuring the influence of their mass on a nano-oscillator. Figure shows a thin rectangular cantilever etched out of silicon (density 2300 kg/m3) with a small gold dot at the end. If pulled down and released, the end of the cantilever vibrates with simple harmonic motion, moving up and down like a diving board after a jump. When bathed with DNA molecules whose ends have been modified to bind with gold, one or more molecules may attach to the gold dot. The addition of their mass causes a very slight-but measurable-decrease in the oscillation frequency. A vibrating cantilever of mass M can be modeled as a block of mass 13M attached to a spring. (The factor of 13 arises from the moment of inertia of a bar pivoted at one end.) Neither the mass nor the spring constant can be determined very accurately-perhaps to only two significant figures-but the oscillation frequency can be measured with very high precision simply by counting the oscillations. In one experiment, the cantilever was initially vibrating at exactly 12 MHz . Attachment of a DNA molecule caused the frequency to decrease by 60 Hz .

Explanation / Answer

F - k x = 0
(m/3) d^2x/dt^2 = k x
d^2x/dt^2 = (3k/m) x = w^2 x
solution is
x = A sin(w t) where w is the angular frequency = 2 pi f
2 pi f = sqrt(3k/m) = (3k)^(1/2) m^(-(1/2))
so 2 pi df = -(1/2) (3k)^(1/2) m^(-(3/2)) dm

dm = -2 m df/f = -2 m (56/12 x 10^6)

will give you the mass of the DNA (the change in mass of the
cantilever m). You didn't give the dimension of the cantilever
so I could not compute m for you, but if you know the dimension
just use the density above to calculate m.
Another more accurate way to approach this is to calculate
the change in the moment of intertia from the added molecule,
but the crudeness of the setup would not seem to warrant that.

'OR'

The spring-mass system vibrates with frequency f, corresponding to an angular velocity =2f, given by:

= sqrt(M/k)
f = /(2) = sqrt(M / (3k)) / (2)

Where k is the spring constant, and the (1/3) inside the square root accounts for the effective mass M/3 of the modeled spring-mass system. This gives two equations:

f = sqrt(M/k) / (2)
f + f = sqrt((M + M)/k) / (2)

We know f and f, but M, M and k are unknown. However, we can estimate M+M to high accuracy as follows.

Take the differential to estimate the effect of a small mass increment:

d = [(1/2) / sqrt(M / (3k))] * dM/ (3k)
d = [(1/2) * sqrt(M / (3k)) * 3k/M] * dM / (3k) = (1/2) * dM/M
d/ = df/f = (1/2) dM/M

So M/M ~= dM/m = 2df/f ~= f/f, to about as many significant digits as we know f/f itself.

This means that M+M is approximately 2*M*(11 MHz + 46 Hz)/(11 MHz). Substituting this above gives two equations with M and k as the only unknowns. Square both equations and solve.

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