Frequency & Wavelength of a Vibrating Wire

  • #1
TRB8985
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Homework Statement
A 5.00 m, 0.732 kg wire is used to support two uniform 235 N posts of equal length. Assume that the wire is essentially horizontal and that the speed of sound is 344 m/s. A strong wind is blowing, causing the wire to vibrate in its 5th overtone. What are the frequency and wavelength this wire produces?
(Theta = 57.0°)
Relevant Equations
f_n = n * v/2L ; v = sqrt(F/mu) ; v_sound = f_n * lambda_sound
Good morning,

I'm working through the problem from the homework statement above and answered it correctly, but I can't help but feel that something important is missing.

I was able to correctly identify the sum of torques by using the following diagram, where the CCW rotational direction represents a positive torque:

UntitledPF3.png


Equating the torque due to the weight of the post and the torque due to the tension (relative to point alpha) looks like this:
$$ W_{post} \cdot \frac {l}{2} cos(\theta) = T \cdot l sin(\theta) $$
This works out great, since the unknown length of the post (which I'm calling ##l##) cancels out and the tension can be used to solve the problem and get the right answer.

However, shouldn't we technically be including an additional torque due to the weight of the wire? Like this?

UntitledPF4.png


I wanted to try this and see what happens, but there's no obvious way to equate the torques which cancels out the length of the post like before:
$$ W_{post} \cdot \frac {l}{2} cos(\theta) = T \cdot l sin(\theta) + W_{wire} \cdot (2.50 m - l cos(\theta)) $$ It's possible to use geometry and determine the internal angles of this isoceles trapezoid figure, but that's not enough information to write the horizontal difference between ##\alpha## and the dashed line in terms of the post's unknown length.

It seems the only option is to just ignore the torque of the wire's weight and hand-wave it away as negligible compared to the other two forces, but that feels rather arbitrary.. is there a better way in the math to handle this situation?

Thank you!
 
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  • #2
Hi,

1. Is ##\theta ## given ?
2. the weight of the wire isn't fully on the attachment point, but uniformly distributed over the 5 m.

Where does the ## 2.50 {\text { m}} - l \cos\theta## come from ?
 
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  • #3
Hey BvU,

Yeah, theta was provided in a picture with this problem. My bad on forgetting to include that. ##\theta = 57.0°##. Added to the homework statement as well.

In regards to your second statement, totally agree. That's why I've placed the center of mass of the wire at its halfway point along its length.

The ## 2.50 m - l cos(\theta) ## comes from the lever-arm between the point of rotation (##\alpha##) and the line of action of the weight of the wire acting at its center of mass:

UntitledPF5.png

Please let me know if anything else is missing or unclear.
 
  • #4
TRB8985 said:
Hey BvU,

Yeah, theta was provided in a picture with this problem. My bad on forgetting to include that. ##\theta = 57.0°##. Added to the homework statement as well.

In regards to your second statement, totally agree. That's why I've placed the center of mass of the wire at its halfway point along its length.

The ## 2.50 m - l cos(\theta) ## comes from the lever-arm between the point of rotation (##\alpha##) and the line of action of the weight of the wire acting at its center of mass:

View attachment 334917
Please let me know if anything else is missing or unclear.

You are freeing the wire from the rod to examine ##T##. You either need to look at things from the perspective of the wire, or the perspective of the rod. I think you are mixing these two perspectives.

EDIT: But maybe that's ok if you can translate ##T## to the dashed line. However, the wire would have to be able to apply a moment to the end of the rod, I guess the question is...can it?

It can't hurt to make a FBD of the wire to see what needs to happen with the forces and moments.
 
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  • #5
TRB8985 said:
Homework Statement: …. Assume that the wire is essentially horizontal

shouldn't we technically be including an additional torque due to the weight of the wire?
Not exactly.
The joint does not 'know' about the length of the wire. It directly experiences the tension.
In reality, the wire will not be quite straight, so the tension will also exert a downward force (the weight of the half wire) on the joint:
## W_{post} \cdot \frac {l}{2} \cos(\theta) +T_y\cdot l \cos(\theta)= T_x \cdot l \sin(\theta) ##
where ##T_y=\frac 12W_{wire}##.
For the frequency, need to consider the whole tension.

Btw, if you prefix trig functions, logs etc. with \ the LaTeX looks a bit cleaner.
 
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Thank you all for the input! I see where I've gone wrong here.
Much appreciated, have a great evening.
 

Related to Frequency & Wavelength of a Vibrating Wire

What is the relationship between the frequency and wavelength of a vibrating wire?

The frequency and wavelength of a vibrating wire are inversely related. As the frequency increases, the wavelength decreases, and vice versa. This relationship is described by the equation \( v = f \lambda \), where \( v \) is the wave speed, \( f \) is the frequency, and \( \lambda \) is the wavelength.

How does the tension in the wire affect its frequency and wavelength?

The tension in the wire directly affects the frequency and wavelength of the vibrations. Increasing the tension in the wire increases the wave speed, which in turn increases the frequency for a given wavelength. Conversely, decreasing the tension lowers the wave speed and frequency.

What factors determine the fundamental frequency of a vibrating wire?

The fundamental frequency of a vibrating wire is determined by the length of the wire, the tension in the wire, and the mass per unit length (linear density) of the wire. The fundamental frequency can be calculated using the formula \( f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( L \) is the length of the wire, \( T \) is the tension, and \( \mu \) is the linear density.

How do harmonics relate to the frequency and wavelength of a vibrating wire?

Harmonics are integer multiples of the fundamental frequency. The nth harmonic has a frequency \( f_n = n \cdot f_1 \) and a wavelength \( \lambda_n = \frac{2L}{n} \), where \( n \) is the harmonic number, \( f_1 \) is the fundamental frequency, and \( L \) is the length of the wire. Each harmonic corresponds to a specific standing wave pattern on the wire.

Can the frequency and wavelength of a vibrating wire be altered without changing its tension or length?

Yes, the frequency and wavelength of a vibrating wire can be altered by changing the boundary conditions or by using external forces such as magnetic fields or electrical currents. For example, adding a mass or damping element to the wire can change its vibrational properties, thus altering the frequency and wavelength.

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