Sound & Music - Mass per unit length

[Torrie]

Homework Statement

Suppose a harp string is tuned to middle C (C4) and is 0.56m long. If I want the tension to be 193.5 N, what mass per unit length do I need the harp string to be? Calculate your answer in kg.m

Homework Equations

Vs = sqrt(T/μ)
V = fλ

The Attempt at a Solution

Having a very hard time. I looked up the frequency of a C4= 261.61 Hz. I calculated Vs as 261.61 x 1.12 = 293. Then I attempted (293 x 193.5)sq, but that number is way too high. I am stuck.

 

[TomHart]

Why did you multiply 293 x 193.5?

 

[Torrie]

I figured if Vs = sqrt (T/μ), then μ= Vs x T sq

 

[TomHart]

If Vs = sqrt(T/μ), then μ = T/V_s², doesn't it?

 

[Torrie]

Thats what I thought initially! But when I answered 0.436 kg/m the it said it was incorrect.

 

[TomHart]

But when I answered 0.436 kg/m the it said it was incorrect.

I don't see how you got 0.436 kg/m. I got a different answer. Can you write out what you did mathematically.

 

[TomHart]

Okay, I see what you did. You divided 193.5 by 293, and then you squared it. It is not (193.5/293)². It is 193.5/(293)².
Or another way to do it is divide 193.5 by 293 and then divide that result by 293 again.

 

[Torrie]

μ = (T/Vs)sq

T = 193.5
Vs = 293

193.5/291 = .66
.664948sq = .4356 kg / m

This answer was not correct

 

[Torrie]

Oh! Okay let me try that

 

[Torrie]

Okay finally! .0022 kg/m is correct. Can you explain to me why it is 193.3/(293) sq instead of (193.3/293)sq?

 

[TomHart]

Okay finally! .0022 kg/m is correct. Can you explain to me why it is 193.3/(293) sq instead of (193.3/293)sq?

Algebraically, that's how the equation turns out when you rearrange it to solve for μ.

V_s = sqrt(T/μ)
Square both sides of the equation to give:
V_s² = T/μ
Multiply both sides of the equation by μ to give:
μV_s² = T
Divide both sides of the equation by V_s² to give:
μ = T/V_s²

Does that make sense?

 

[Torrie]

That does make sense! Thank you so much for all of your help. I really appreciate it!