Mastering Physics Solutions: Exercise 14.52

Mastering Physics Solutions: Exercise 14.52

On December 20, 2011, in Chapter 14: Sound, by Mastering Physics Solutions

Exercise 14.52

Part A = fb = 4.2 Hz Click to use the calculator/solver for this part of the problem

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?

Express your answer using two significant figures.

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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