(Q1) The figure shows a 180 g uniform rod pivoted at one end. The other end is a

Question

(Q1)

The figure shows a 180 g uniform rod pivoted at one end. The other end is attached to a horizontal spring. The spring is neither stretched nor compressed when the rod hangs straight down.
What is the rod's oscillation period? You can assume that the rod's angle from vertical is always small.

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(Q2)

A spring with spring constant 15.5 N/m hangs from the ceiling. A 590 g ball is attached to the spring and allowed to come to rest. It is then pulled down 6.70 cm and released.
What is the time constant if the ball's amplitude has decreased to 4.00 cm after 45.0 oscillations?

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(Q3)

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/m 3 ) 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 1 3 M attached to a spring. (The factor of 1 3 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 11 MHz . Attachment of a DNA molecule caused the frequency to decrease by 55 Hz .
What was the mass of the DNA?

Thank you :)

Explanation / Answer

Q1) when rod is displaced by @ angle form vertical,

torque on rod = (L/2)x mgsin@ + k(Lsin@) = (mg + k) L sin@

when @ theta is smalle

F = (mg +k ) L @

time period = 2pi sqrt[ (m L^2/3) / (mg +k)L)

T = 2pi sqrt (L / 3(mg + k))

L = 0.20 m , m = 0.180 kg , k = 3 N/m

T = 0.743 sec

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