How Do Changes in Tension Affect the Frequency of Musical Instruments?

[boilerpilot]

Homework Statement

The tension in a guitar string is 21% too high. The fundamental frequency will be...

Homework Equations

V=F$$\lambda$$=$$\sqrt{Tension}$$

The Attempt at a Solution

I know the answer is 10% too high, but not sure how to get there...
If I take the square root of 1.21, I get 1.1 (which would equal the velocity) but without lamba, I cannot find the frequency...

A similar problem...

Homework Statement

A piano tuner finds that the frequency of a note is 417Hz instead of 396Hz. By what percentage should the tension in the string be changed to correct this?

Homework Equations

Same equations as above...

The Attempt at a Solution

Not sure on this one either...the ratio of the frequencies is 1.053. I'm sure that comes into play, but not sure how to solve. The answer is 10% on this one as well...

 

[anathema]

it is important to approach problems like these using mathematical equations and logic. In the first problem, you are given the tension in the guitar string and asked to find the corresponding fundamental frequency. The equation for fundamental frequency is V=Fλ, where V is the velocity of the wave, F is the frequency, and λ is the wavelength. In this case, you are given the tension (T) and asked to find the frequency (F). So, rearranging the equation, we get F=√(T/μ), where μ is the linear density of the string. Since the tension is 21% too high, we can write T'=T*1.21, where T' is the new tension. Substituting this into the equation, we get F'=√(1.21T/μ). Since we want to find the percentage by which the frequency is too high, we can divide F' by the original frequency (F) and subtract 1. So, the percentage is (F'/F-1)*100. Substituting the values, we get (√(1.21T/μ)/√(T/μ)-1)*100. Simplifying this, we get (1.1-1)*100=10%. So, the fundamental frequency of the guitar string is 10% too high.

In the second problem, you are given the frequency of a note and asked to find the percentage change in tension needed to correct it. Again, we can use the equation F=√(T/μ). Since the frequency is 417Hz instead of 396Hz, we can write F'=417Hz and F=396Hz. Substituting these values into the equation, we get 417Hz=√(T'/μ) and 396Hz=√(T/μ). Dividing the first equation by the second, we get (417/396)=√(T'/T). Squaring both sides, we get (417/396)^2=T'/T. Simplifying this, we get T'=1.053T. This means that the new tension (T') should be 1.053 times the original tension (T). So, the percentage change in tension is (T'/T-1)*100=(1.053-1)*100=10%. Therefore, the tension in the string should be increased by 10%