Question

the frequency of a stringed instrument is inversely proportional to the string length. if an instrument produces 602 hertz frequency with a string length of 60 cm, then what will the frequency be for 33 string length in cm?
hint: remember an inverse proportion means the ratio of either frequencies or string lengths needs be reversed or inverted.
$y = \frac{k}{x}$ where y can represent the hertz and x can represent the string, while k is a constant.
the frequency is in hertz, but you do not need to use a label for your answer. you only need to enter the number. round to the nearest tenth if necessary.
review the linked section of the prealgebra textbook for problem help

Explanation:

Step1: Find the constant \( k \)

Given \( y=\frac{k}{x} \), when \( y = 602 \) and \( x = 60 \), we substitute into the formula:
\( 602=\frac{k}{60} \)
Multiply both sides by 60 to solve for \( k \):
\( k=602\times60 = 36120 \)

Step2: Find the new frequency \( y \)

Now we have \( k = 36120 \) and \( x = 33 \). Substitute into \( y=\frac{k}{x} \):
\( y=\frac{36120}{33}\)
Calculate \( \frac{36120}{33}= 1094.5454\cdots\)

Answer:

\( 1094.5 \)