Beat frequency for hanging bar between two wires with added offset weight

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Homework Help Overview

The problem involves a uniform bar supported by two wires, with an additional weight affecting the tension in the wires. Participants are exploring the calculation of beat frequencies produced when the wires are plucked, focusing on the fundamental frequency equations and the impact of the weight's position on tension.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss calculating the frequencies of two wires based on their tensions, questioning the distribution of tension due to the added weight. There is also a focus on the implications of beat frequencies and their audibility.

Discussion Status

Some participants have confirmed numerical values for tension and beat frequency, while others express confusion regarding the relationship between beat frequency and audibility. The conversation is ongoing, with attempts to clarify misunderstandings about beat frequencies.

Contextual Notes

Participants are navigating assumptions about how tension is affected by the placement of the weight and the implications of frequency differences on auditory perception. There is mention of a teacher's guidance that may influence understanding.

NickP89
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Homework Statement


A uniform 165 N bar is supported horizontally by two identical wires A and B. A small 185 N cube of lead is placed 3/4 of the way from A to B. The wires are each o.75 m long and have a mass of 5.50 g. If both of them are simultaneously plucked at the center, what is the frequency of the beats that they will produce when vibrating in their fundamental?

Homework Equations


fundamental frequency = (1/2L) * √(F/μ).
where L is the length of the wire, (μ) is the mass per unit length, F is the tension in the wire
f(beat) = f1-f2.

The Attempt at a Solution


I tried to calculate the frequency of each wire A and B.
I let frequency1 be the frequency of the wire with the block closer to it, so wire B.

frequency1=(1/(2*0.75m))*√(F1)/(0.0055kg/0.75m)
frequency1=7.785*√F1

frequency2=7.785*√F2, as L and μ are the same for both wires.

This is where I got stuck. I tried to calculate the tensions:

F1= (1/2)*(165N) + (3/4)*185 = 221.25N
F2= (1/2)*(165N) + (1/4)*185 = 128.75N

I'm not sure if the tension due to the block would be distributed in the way I did.
When i put these tensions into the equation i get a large number, but i know that beats don't occur at frequency differences greater than 10-15 Hz.

Thank you.
 
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I have not checked your numerical answers all the way through but I do get the tension to be the same as yours.
I can't see anything wrong with your method.
I got the beat freq1uency to be 28Hz.
I will double check my numbers and correc this if it is wrong !
 
It is 28 Hz. I was a bit confused when i got this answer because my teacher said you don't hear beats at frequency differences greater than 15 Hz. Thanks for your help
 
Beat frequency is a frequency in its own right! It is the difference between the 2 frequencies and if that is an audio frequency you will hear it.
Do you play a musical instrument? Beat frequencies are used to tune instruments.
 
Sorry what i meant was you don't hear the individual beats and that it merges into dissonance or consonance. I should have just gone through with the 28 Hz. Thank you though, now i understand beat frequencies do occur at greater values. And no i don't play any instruments but my prof demonstrated tuning a guitar.
 
I know what you mean... low beat frequencies sound like a 'throb' rather than a 'note'
Your prof probably got to produce slow throbs as he tuned his guitar!
But the beat frequency can sound (is!) a genuine note in its own right.
Nice thread this...
 

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