How Does Wind Affect the Frequency of a Vibrating Wire in Its 7th Overtone?

  • Thread starter Thread starter rteng
  • Start date Start date
  • Tags Tags
    Frequency
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a vibrating wire subjected to wind, specifically focusing on the frequency of the sound produced in its 7th overtone. The wire's parameters include its length, mass, and the weight of posts it supports, with additional factors like tension and torque being considered in the analysis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of tension in the wire and its relationship to the forces acting on the posts. There are attempts to balance forces and torques, with some questioning the assumptions about the forces involved and the method for calculating torque without knowing the length of the pole.

Discussion Status

There is ongoing exploration of the correct approach to calculating tension and torque, with various interpretations being discussed. Some participants have provided guidance on using torque equations and the significance of the forces acting at specific points, while others are still seeking clarity on the calculations and assumptions made.

Contextual Notes

Participants note the challenge of calculating torque without specific lengths and the need to clarify the roles of different forces in the system. The weight of the posts and its effect on tension are also under scrutiny, indicating a complex interplay of forces that may not be fully resolved yet.

rteng
Messages
26
Reaction score
0

Homework Statement



5.10m, 0.733kg wire is used to support two uniform 230N posts of equal length. Assume that the wire is essentially horizontal and that the speed of sound is 344m/s. A strong wind is blowing, causing the wire to vibrate in its 7th overtone.

YF-15-37.jpg

What is the frequency of the sound this wire produces?

Homework Equations



fn=nv/2L
v=sqrt(F/linearmass)
sumF=0

The Attempt at a Solution



-split the force vectors into their components
-multiplied the force along the x-axis from one of the posts by two to find the tension in the string
-used this value to find velocity
-used velocity to find the frequency of the 7th overtone
-got answer wrong

help?
 
Physics news on Phys.org
What tension did you get in the string?
 
Balancing torques will be correct, I suppose.
 
T=250.5n
 
rteng said:
T=250.5n

That's not what I get. How did you get that?
 
T=250.5
v=sqrt(250.5/(0.733/5.1))
v=41.75
f(6)=6(v/2L)
f(6)=24.6

thats my initial work
however it is incorrect

what was done wrong?
 
230cos(57)=125.3

125.3X2=250.5
 
Tension is not balancing forces. Its their torques which are balancing.
 
rteng said:
230cos(57)=125.3

125.3X2=250.5

This isn't correct. The 230N isn't the tension/compression in the post... it is just the weight of the post. As Sourabh N said, use torque about one of the pivots (either one doesn't matter which one)... net torque about the pivot is 0.
 
  • #10
ok I understand to use torque
but how can I find the torque without the length of the pole?
maybe I am unaware of a method to calculate torque
 
  • #11
rteng said:
ok I understand to use torque
but how can I find the torque without the length of the pole?
maybe I am unaware of a method to calculate torque

The length will cancel... write out the equation for torque about the pivot...
 
  • #12
torque= rXF

I don't seem to understand how to use this formula without knowing the position of the forces.
is the force in this equation acting on the string?
where are the 230N of the pole acting?
 
  • #13
rteng said:
torque= rXF

I don't seem to understand how to use this formula without knowing the position of the forces.
is the force in this equation acting on the string?
where are the 230N of the pole acting?

The 230N force is acting at the center of the post (midpoint)... The tension is acting at the end of the post...
 
  • #14
ok so the tension balances the torque.

and I can find torque with the formula torque=rXF

and F is 230N down

the position of this force is at the midpoint of the post

therefore, I can't get a numerical value for torque...

r(pivot to force)=0.5(length of pole)?
 
  • #15
Torque is r X F. This can be interpreted as force times the perpendicular distance from the line of force to the pivot... ie the torque due to the weight is 230N*(L/2)cos57... if we take clockwise positive and counterclockwise negative and look at the left pivot... torque due to the weight is -230*(L/2)cos57

What is the torque due to tension about the left pivot? The force is T. What is the perpendicular distance from the line of force to the pivot?

The sum of these 2 torques is 0.
 
  • #16
the torque on the right pivot is 115(L)cos57
the torque on the left pivot is -115(L)cos57

so if I use torque=Fd

and the torque is the above calculated values...and the force is the tension in the wire and d=(L)sin57

then F(right pivot)= (115(L)cos57)/(sin57(L))
F=74.68N

so the total tension is 74.68X2=149.36N

the use this value for the tension in the string

is that correct?
 
  • #17
You don't need to multiply by 2. The tension is just 74.68N.
 
  • #18
T=74.68N

v=sqrt(74.68/(0.733/5.1))
v=22.79m/s

f=6(v/2L)
f=13.4Hz

what is wrong with those calculations?

the answer is still wrong
 
  • #19
I think you should be multiplying by 8 not 6.
 
  • #20
oops silly mistake

thank you
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
11K
Replies
6
Views
2K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
3
Views
2K