Chapter 8, Problem 1CA

To determine

Find the approximate frequency of label $4$ using diagram.

Answer to Problem 1CA

The right option is $B$ .

Explanation of Solution

Given:

The given diagram is

  

The frequencies of the notes form a geometric sequence.

Calculation:

The general rule for the $n\text{th}$ term of the geometric sequence is $a_n=a_1{r}^{n-1}$ , where $a_1$ is first term and $r$ is the common ratio.

In the given diagram

  $\begin{array}{l}{a}_8=a_1{r}^{8-1}=392\\ \Rightarrow a_1{r}^7=392---\left(i\right)\\ a_{10}=a_1{r}^{10-1}=440\\ \Rightarrow a_1{r}^9=440---\left(ii\right)\end{array}$

Find the common ratio.Divide $\left(ii\right)$ by $\left(i\right)$ .

  $\begin{array}{l}\frac{a_1{r}^9}{a_1{r}^7}=\frac{440}{392}\\ \Rightarrow r^{9-7}=\frac{55}{49}\\ \Rightarrow r^2=\frac{55}{49}\\ \Rightarrow r=\sqrt{\frac{55}{49}}\\ \Rightarrow r=1.059\end{array}$

Substitute the value of $r=1.059$ in equation $\left(i\right)$ .

  $\begin{array}{l}{a}_1{r}^7=392\\ \Rightarrow a_1=\frac{392}{r^7}\end{array}$

Now,

  $\begin{array}{l}{a}_4=a_1{r}^{4-1}=a_1{r}^3\\ \Rightarrow a_4=\frac{392}{r^7}{r}^3\\ \Rightarrow a_4=\frac{392}{r^4}\\ \Rightarrow a_4=\frac{392}{{\left(1.059\right)}^4}\\ \Rightarrow a_4=311.67\text{Hz}\end{array}$

Hence the right optionis $B$ .