Exercise 14.52

Solutions Below:

Part A

Express your answer using two significant figures.

The beat is just the difference between the two frequencies (f_{new} – f_{old}). Start by finding the new frequency.

We know that:

(Τ_{new} – Τ_{old}) / (Τ_{old}) = 0.019

And since Τ = μv^{2}:

Τ = μv^{2}

(μv_{new}^{2} – μv_{old}^{2}) / μv_{old}^{2} = 0.019

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

(μv_{new}^{2} – μv_{old}^{2}) / μv_{old}^{2} = 0.019

and

v = λf:

(μ(λf_{new})^{2} – μ(λf_{old})^{2}) / μ(λf_{old})^{2} = 0.019

The μ and λ divide out:

(f_{new}^{2} – f_{old}^{2}) / f_{old}^{2} = 0.019

Solve for f_{new} accordingly:

(f_{new}^{2} – f_{old}^{2}) / f_{old}^{2} = 0.019

(f_{new}^{2} – f_{old}^{2}) = 0.019f_{old}^{2}

f_{new}^{2} = 0.019f_{old}^{2} + f_{old}^{2}

f_{new} = sqrt(0.019f_{old}^{2} + f_{old}^{2})

f_{new} = sqrt(0.019 * 440^{2} + 440^{2})

f_{new} = sqrt(197278.4)

f_{new} = 444.16Hz

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

f_{beat} = f_{new} – f_{old}

f_{beat} = 444.16 – 440

f_{beat} = 4.2Hz (2 significant figures)