Chapter 7.3, Problem 85PPS

a.

To determine

To calculate the frequency of concert A.

Answer to Problem 85PPS

440 Hz

Explanation of Solution

Given:

Expression: $f=440{(2^{\frac{1}{12}})}^{n-49}$

  

Formula used:

  ${(a^b)}^c=a^{bc}$

  $a^0=1$

Calculation:

Consider the expression that represents the frequency.

  $f=440{(2^{\frac{1}{12}})}^{n-49}$ where n represents nth key on the piano.

From the figure,

In concert A, nth key represents the 49th key.

Substituting the value of nth key as 49,

  $\begin{array}{c}$ f=440{(2^{\frac{1}{12}})}^{n-49}$$ =440{(2^{\frac{1}{12}})}^{49-49}$$ =440{(2^{\frac{1}{12}})}^0$=440(1)\\ =440\end{array}$

Conclusion:

Therefore, the frequency of concert A is 440 Hz.

b.

To determine

To find the key that has the frequency 220 Hz.

Answer to Problem 85PPS

37^th key

Explanation of Solution

Given:

Frequency: 220 Hz

Formula used:

  ${(a^b)}^c=a^{bc}$

  $a^0=1$

Calculation:

Consider the expression that represents the frequency,

  $f=440{(2^{\frac{1}{12}})}^{n-49}$

Substituting the value of frequency 220 Hz.

  $f=440{(2^{\frac{1}{12}})}^{n-49}$

  $220=440{(2^{\frac{1}{12}})}^{n-49}$

Dividing 440 on both sides,

  $\begin{array}{c}$ \frac{220}{440}=\frac{440}{440}{(2^{\frac{1}{12}})}^{n-49}$$ \frac{1}{2}={(2^{\frac{1}{12}})}^{n-49}$$ 2^{-1}={(2^{\frac{1}{12}})}^{n-49}[\because \frac{1}{a^b}=a^{-b}]$-1=\frac{1}{12}(n-49)[\because a^{f(x)}=a^{g(x)},\text{then }f(x)=g(x)]\\ n-49=-12\\ n=49-12\\ n=37\end{array}$

Conclusion:

Therefore, the 37^th key has frequency 220 Hz.