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- Thread starter Jimmy87
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Quantum Defect

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Yup. You have a vibrating bar, clamped at both ends. The lowest frequency oscillation of the ruler/bar will have one half a wavelength. Moving the supports inwards shortens the wavelength. This assumes very small displacements of the ruler. What might happen if the displacement becomes larger?

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sophiecentaur

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If you want to see the details of the way the sinusoid is obtained from the simple (Hooke's Law) force law of a spring acting on a mass, then see this link.

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If you want to see the details of the way the sinusoid is obtained from the simple (Hooke's Law) force law of a spring acting on a mass, then see this link.

Thanks guys. When you move the supports closer, the displacement/depression (distance at the center between unloaded and loaded) decreases. Why is this? I was thinking you could model this with hooke's law saying that the spring constant k, increases as the supports are moved in. But what is the exact reason for the stiffness going up?

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Nugatory

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Why is this? I was thinking you could model this with hooke's law saying that the spring constant k, increases as the supports are moved in. But what is the exact reason for the stiffness going up?

If the supports are closer together, you have to bend the rod more to displace the center by the same amount. To bend it more, you need more force, so you're using more force for that same displacement - that's in effect an increase in teh spring constant.

(To see this, imagine that the bent rod assumes the shape of a circular arc passing through the two supports and the displaced position - the circle will be smaller and hence more sharply curved as the supports are moved closer together if you keep the displacement constant. The actual shape is not circular arc, but it's close enough to provide an OK qualitative picture).

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Quantum Defect

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This is identical to what is going on in xylophones, marimbas, etc. There are usually nice derivations of the physics of these systems in Musical Acoustics books. Look at Benade's book (easier) as well as Fletcher and Rossing (gnarlier).Thanks guys. When you move the supports closer, the displacement/depression (distance at the center between unloaded and loaded) decreases. Why is this? I was thinking you could model this with hooke's law saying that the spring constant k, increases as the supports are moved in. But what is the exact reason for the stiffness going up?

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sophiecentaur

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Without doing the sums, you could think of a simple spring. It requires twice the force to extend a long spring by a given distance at its half way point that it requires for the full length of spring.But what is the exact reason for the stiffness going up?

There is a long wiki article about bending beams (rulers, in your case).

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Chestermiller

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If the mass of the beam is negligible compared to mass of the weight, then wave effects are insignificant. The beam deflection at any time will be the same as if it were under static load equal to the instantaneous tension in the wire. The relation between the tension in the wire and the deflection will be determined by the static elastic beam bending behavior.Yup. You have a vibrating bar, clamped at both ends. The lowest frequency oscillation of the ruler/bar will have one half a wavelength. Moving the supports inwards shortens the wavelength. This assumes very small displacements of the ruler. What might happen if the displacement becomes larger?

Chet

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Chestermiller

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For a

[tex]δ=\frac{l^3}{48EI}F[/tex]

where l is the distance between the supports, I is the moment of inertia of the beam about its neutral axis, and E is the Young's modulus of the beam material. So the force-displacement equation is given by:

[tex]F=\frac{48EI}{l^3}δ[/tex]

and the "spring constant k" for the deflection is given by:

[tex]k=\frac{F}{δ}=\frac{48EI}{l^3}[/tex]

The deflection is a parabolic function of distance along the beam (and not sinusoidal).

So tell me, where are the "waves" in this

If you hang a weight of mass M in the middle of a beam of negligible mass (compared to M), and subject it to simple harmonic motion, the frequency of the up-down oscillation will be given by:

[tex]2πf=\sqrt{\frac{k}{m}}=\sqrt{\frac{48EI}{Ml^3}}[/tex]

At any instant of time, the deflection of the beam will be parabolic (not sinusoidal).

The frequency of the oscillation will be inversely proportional to the distance between supports l to the 3/2 power.

So now I've shown you my solution to this problem. Now it's time for you to show us your wave solution to this same problem (under the constraint that the mass of the beam is negligible compared to the mass M hanging from the beam). Also tell us how, in ordinary simple harmonic motion of a mass and massless spring, how the waves and the speed of sound come in.

Chet

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Quantum Defect

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I was thinking of something along the lines of this:If the mass of the beam is negligible compared to mass of the weight, then wave effects are insignificant. The beam deflection at any time will be the same as if it were under static load equal to the instantaneous tension in the wire. The relation between the tension in the wire and the deflection will be determined by the static elastic beam bending behavior.

Chet

With the lowest mode vibration being excited by "plucking" the mass hanging from the ruler. Decreasing the spacing between the supports, decreases the wavelength, increases the frequency. I failed to take into account the wiggling of the ends (you will have two cantilevers hanging off the two supports on the ends) , which makes my simple-minded picture less than perfect -- but the frequency should go up as the supports are moved inwards.

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Chestermiller

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Thanks. This all applies to a beam that has mass, but and has no weight hanging from it. That's pretty different from the case the OP was interested in, which has a substantial weight dangling from the center. The result that I gave assumes that the dangling mass dominates, and that the mass of the beam is negligible compared to the dangling mass. In that case, there is no wave solution. Also, the case I considered is not cantilevered at the ends. The ends are simply supported, meaning that the bending moments on the ends are zero.I was thinking of something along the lines of this:

With the lowest mode vibration being excited by "plucking" the mass hanging from the ruler. Decreasing the spacing between the supports, decreases the wavelength, increases the frequency. I failed to take into account the wiggling of the ends (you will have two cantilevers hanging off the two supports on the ends) , which makes my simple-minded picture less than perfect -- but the frequency should go up as the supports are moved inwards.

Chet

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