## Mastering Physics Solutions: Exercise 14.52

Exercise 14.52

Part A = fb = 4.2 Hz

Solutions Below:

Two identical strings on different cellos are tuned to the 440 Hz A note. The peg holding one of the strings slips, so its tension is decreased by 1.9 %.

Part A

What is the beat frequency heard when the strings are then played together?

The beat is just the difference between the two frequencies (fnew – fold). Start by finding the new frequency.

We know that:

new – Τold) / (Τold) = 0.019

And since Τ = μv2:

Τ = μv2
(μvnew2 – μvold2) / μvold2 = 0.019

since v = λf and the problem says that λ stays the same:

(μvnew2 – μvold2) / μvold2 = 0.019

and

v = λf:

(μ(λfnew)2 – μ(λfold)2) / μ(λfold)2 = 0.019

The μ and λ divide out:

(fnew2 – fold2) / fold2 = 0.019

Solve for fnew accordingly:

(fnew2 – fold2) / fold2 = 0.019
(fnew2 – fold2) = 0.019fold2
fnew2 = 0.019fold2 + fold2
fnew = sqrt(0.019fold2 + fold2)
fnew = sqrt(0.019 * 4402 + 4402)
fnew = sqrt(197278.4)
fnew = 444.16Hz

Now that we have the new frequency, we can find the beat:

fbeat = fnew – fold
fbeat = 444.16 – 440

fbeat = 4.2Hz (2 significant figures)

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