# The frequency of sound emitted by a string is inversely proportional to the length of the string. When the string is 50 cm long, the frequency of sound is 256Hz. Find (a) the frequency of sound when the string is 40 cm long (a) the length of the string when the frequency of sound is 400Hz

We are told that the frequency of sound emitted by a string is inversely proportional to its length. Also, the frequency is 256Hz when the length is 50cm.

Note that when variables vary inversely, as one variable increases the other variable decreases and vice versa.

When two variables are inversely proportional (or vary inversely) the variables are related in the following manner: `y=k/x ` where the variables are x,y and k is called the constant of proportionality. Here we have `f=k/l ` where f is the frequency in hertz, l is the length in centimeters, and k is the constant of proportionality. Note that k will depend on outside factors such as the material of the string, the medium it is in, etc..., but k will be constant for the given situation.

We are given that when f=256, l=50, so we can solve for k:

`256=k/50 ==> k=50(256)=12800 `

(a) If l=40 we have `f=12800/40 ==> f=320 ` ; so when the length is 40cm the frequency is 320Hz.

This agrees with the knowledge that as the length gets shorter the frequency should increase.

(b) If f=400 then `400=12800/l ==> l=12800/400 ==> l=32 ` ; thus when the frequency is 400Hz, the length will be 32cm.

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We are told that the frequency of the sound emitted by the string is inversely proportional to the length of the string, and that a 50 cm string emits 256 Hz sound.

We are not told all the details of the equation relating string length and sound. But we have data to use to solve for the unknown portion.

`f = X / L`

Where X is the unknown portion of the equation. We know that L is on the bottom because an inverse relationship between y and z means

`y=1/z`

By plugging in the known data (50 cm and 256 Hz), we can solve for X.

`X=f*L=50*256=12800`

Now we are able to use the following equation to solve parts a and b.

`f=12800/L`

A. `f=12800/40 = 320 Hz`

B. `L = (12800/f) = (12800/400) = 32`

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