Frequency of the 7th Overtone

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In summary: I think I understand what I need to do now.In summary, the wire in a strong wind produces a vibratory sound at a frequency of 13.4Hz.
  • #1
rteng
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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?
 
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  • #2
What tension did you get in the string?
 
  • #3
Balancing torques will be correct, I suppose.
 
  • #4
T=250.5n
 
  • #5
rteng said:
T=250.5n

That's not what I get. How did you get that?
 
  • #6
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?
 
  • #7
230cos(57)=125.3

125.3X2=250.5
 
  • #8
Tension is not balancing forces. Its their torques which are balancing.
 
  • #9
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
 

1. What is the frequency of the 7th overtone?

The frequency of the 7th overtone is seven times the fundamental frequency of the vibrating object.

2. How is the frequency of the 7th overtone calculated?

The frequency of the 7th overtone can be calculated by multiplying the fundamental frequency by 7.

3. What is the relationship between the frequency of the 7th overtone and the fundamental frequency?

The frequency of the 7th overtone is directly proportional to the fundamental frequency. This means that as the fundamental frequency increases, the frequency of the 7th overtone will also increase.

4. What is the significance of the 7th overtone in sound waves?

The 7th overtone is an important component in the harmonic series of a sound wave. It contributes to the overall timbre or quality of the sound produced by an object.

5. How does the amplitude of the 7th overtone compare to the fundamental frequency?

The amplitude of the 7th overtone is typically lower than that of the fundamental frequency. This means that the 7th overtone will be quieter and have a less significant impact on the overall sound produced by an object.

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