Simple Harmonic Motion experiment

In summary, the conversation discusses an experiment involving a ruler and a weight that undergoes simple harmonic motion (SHM) when pulled down. It is observed that moving the supports closer together increases the frequency of oscillation. The physics behind this is explained as the effective wavelength of the standing wave becoming smaller, causing the frequency to increase to maintain a constant speed of propagation. The stiffness of the system is also discussed, with the closer supports requiring more force to bend the ruler, resulting in a higher spring constant. The relevance of this experiment to musical instruments such as xylophones and marimbas is also mentioned.
  • #1
Jimmy87
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Hi pf. I have a question about an experiment. If you lie a ruler horizontally across two supports (one near each end) and then hang a weight in the middle it will undergo SHM if you pull the mass down. If you move the supports in closer it will oscillate with a higher frequency. I just wanted to know the physics behind this. The only thing I can think of is modelling it as a standing wave. As you move the supports in the effective wavelength of the standing wave is smaller so the frequency must increase to keep the speed of propagation the same. Is that correct?
 
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  • #2
Jimmy87 said:
Hi pf. I have a question about an experiment. If you lie a ruler horizontally across two supports (one near each end) and then hang a weight in the middle it will undergo SHM if you pull the mass down. If you move the supports in closer it will oscillate with a higher frequency. I just wanted to know the physics behind this. The only thing I can think of is modelling it as a standing wave. As you move the supports in the effective wavelength of the standing wave is smaller so the frequency must increase to keep the speed of propagation the same. Is that correct?
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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  • #3
If you write down the equation of motion of the weight - (ideally) the restoring force will be proportional to the distance that the ruler is displaced from the equilibrium position and this force will produce an acceleration on the mass (treating the whole thing as a single mass). This will give you a second order differential equation which, when solved, will give you a sinusoidal variation of the position with time.
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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  • #4
sophiecentaur said:
If you write down the equation of motion of the weight - (ideally) the restoring force will be proportional to the distance that the ruler is displaced from the equilibrium position and this force will produce an acceleration on the mass (treating the whole thing as a single mass). This will give you a second order differential equation which, when solved, will give you a sinusoidal variation of the position with time.
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?
 
  • #5
Jimmy87 said:
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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  • #6
Jimmy87 said:
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?
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).
 
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  • #7
Jimmy87 said:
But what is the exact reason for the stiffness going up?
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.
There is a long wiki article about bending beams (rulers, in your case).
 
  • #8
Quantum Defect said:
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?
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
 
  • #9
I think Jimmy's original suggestion is spot on. Small amplitude motions are carried by sound waves traveling in the material. In a rod, this speed is mostly independent of length. Therefore, the resonance frequencies depend on wavelength set by the distance between supports.
 
  • #10
Khashishi said:
I think Jimmy's original suggestion is spot on. Small amplitude motions are carried by sound waves traveling in the material. In a rod, this speed is mostly independent of length. Therefore, the resonance frequencies depend on wavelength set by the distance between supports.
For a statically loaded simply supported beam with a downward force F applied at its center, the downward displacement of the center of the beam δ is given by the equation:
[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 static equilibrium equation?

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
 
  • #11
Chestermiller said:
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
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.
 
  • #12
Quantum Defect said:
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.

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.

Chet
 

1. What is Simple Harmonic Motion (SHM)?

Simple Harmonic Motion is a type of periodic motion where an object moves back and forth along a straight line, with a constant amplitude and a restoring force that is directly proportional to its displacement from the equilibrium position.

2. How can I demonstrate Simple Harmonic Motion in an experiment?

One way to demonstrate SHM is by using a simple pendulum, where a mass is suspended from a string and allowed to swing back and forth. Other experiments can also be used, such as a mass on a horizontal spring or a mass on a vertical spring.

3. What are the factors that affect Simple Harmonic Motion?

The factors that affect SHM include the mass of the object, the spring constant, and the amplitude of the motion. The greater the mass or spring constant, the slower the oscillation, while a larger amplitude will result in a faster oscillation.

4. How do I calculate the period and frequency of Simple Harmonic Motion?

The period of SHM is the time it takes for one complete cycle or oscillation, while the frequency is the number of cycles per unit time. The period can be calculated using the equation T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant. The frequency can be calculated by taking the inverse of the period (f = 1/T).

5. What are the applications of Simple Harmonic Motion in real life?

Simple Harmonic Motion has many real-life applications, such as in clocks, pendulum clocks, and musical instruments. It is also used in the design of shock absorbers, car suspensions, and earthquake-resistant buildings. Understanding SHM is crucial in many fields of science and engineering, including physics, mechanics, and acoustics.

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