Homework Help: Frequency of the 7th Overtone

1. Oct 9, 2007

rteng

1. The problem statement, all variables and given/known data

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.

What is the frequency of the sound this wire produces?

2. Relevant equations

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

3. 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

help?

2. Oct 9, 2007

learningphysics

What tension did you get in the string?

3. Oct 9, 2007

Sourabh N

Balancing torques will be correct, I suppose.

4. Oct 9, 2007

T=250.5n

5. Oct 9, 2007

learningphysics

That's not what I get. How did you get that?

6. Oct 9, 2007

rteng

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. Oct 9, 2007

rteng

230cos(57)=125.3

125.3X2=250.5

8. Oct 9, 2007

Sourabh N

Tension is not balancing forces. Its their torques which are balancing.

9. Oct 9, 2007

learningphysics

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. Oct 9, 2007

rteng

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. Oct 9, 2007

learningphysics

The length will cancel... write out the equation for torque about the pivot...

12. Oct 9, 2007

rteng

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. Oct 9, 2007

learningphysics

The 230N force is acting at the center of the post (midpoint)... The tension is acting at the end of the post...

14. Oct 9, 2007

rteng

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. Oct 9, 2007

learningphysics

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. Oct 9, 2007

rteng

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. Oct 9, 2007

learningphysics

You don't need to multiply by 2. The tension is just 74.68N.

18. Oct 9, 2007

rteng

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?

19. Oct 9, 2007

learningphysics

I think you should be multiplying by 8 not 6.

20. Oct 9, 2007

rteng

oops silly mistake

thank you