Hi Ian!

We can start with two of the related equations you probably have used before:

f_{n} = nv/2L for the harmonic frequencies of a string fixed at both ends

where

n = order number

v = speed of wave on the string

L = length of string

And:

v = √(T/μ)

where

T = tension in the string

μ = mass density (= m/L if the string is uniform, which is usually assumed)

You can combine these into:

f_{n} = (n/2L)√(T/μ)

Now, we are not told which harmonic of the string this is, but strings usually vibrate with the most power in their fundamental harmonic, so we can pretty safely take n = 1. We can then solve for T_{1}, the tension in the string tuned to 270 Hz.

270 Hz = (1/[2(0.67 m)])√(T_{1}/(9.0x10^{-4} kg/m))

This can be solved for T_{1.}

You can also determine the frequency of the other string, but you need a bit more information, since the realtionship between beat frequency and the two frequencies that compose it is:

f_{beat} = |f_{1}-f_{2}|

We know one of the frequencies in 270 Hz, but the other could be EITHER 250 Hz or 290 Hz to produce a 20-Hz beat frequency. And which one it is will factor into the answer to the question. So we have to determine which of those possibilities the second frequency is. We can then determine the tension in the string to be tuned, and therefore know how much it has to be changed to match the tension, already calculated, for the 270-Hz string.

We are told the beat frequency goes up when the tension in the second string is increased. By the equation we wrote above:

f_{n} = (n/2L)√(T/μ)

increased tension T results in an increased frequency. If the beat frequency goes up, that means the difference between the two frequencies is getting larger. Thus, increasing the second string's frequency moves it further from the frequency of the first string. This means it has to start at f = 290 Hz.

By the same method as above, you can calculate the tension needed to achieve f = 290 Hz. Then you can compute the change in tension needed to get from f = 290 Hz to f = 270 Hz.

I hope this helps get you on your way. If you have any questions about this, or would like to check an answer, please let me know.

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