Question 472321: On mindy's guitar, the frequency of a vibrating string varies inversely as the length of the sting. When a guitar sting of s inches in length of the string. When a guitar sting of s inches in length that vibrates at a frequency of 80 cycles per second is shortened to 6 inches in length, it vibrates at a frequency of 120 cycles per second. What is the value of s?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the formula for a direct ratio is s1/s2 = f1/f2.
the formula for an inverse ratio is s1/s2 = f2/f1.
s is the string length.
f if the frequency that the string vibrates at.
when s1 = x, f1 = 80
when s2 = 6, f2 = 120
we'll use the inverse ratio:
s1/s2 = f2/f1 becomes:
x/6 = 120/80
cross multiply to get:
80*x = 6*120
divide both sides by 80 to get:
x = 6*120/80 which becomes:
x = 9
our data table becomes:
when s1 = 9, f1 = 80
when s2 = 6, f2 = 120
this is an inverse relationship so we're good.
we can also solve it using the k formula of:
direct relationship is y = k*x.
inverse relationship is y = k/x.
we'll use the inverse relationship formula.
first we solve for k and then use that to solve the problem.
we let x = length of the string and we let y = the vibration in cycles per second.
given that when the length is 6, the vibrations are 120 cycles per second, we plug that into the equation and we get:
120 = k/6
solve for k to get:
k = 6*120 = 720
use that value of k to solve the problem.
when the length of the string is x, the vibrations are 80 cycles per second.
plug that into the equation and make k = 720 to get:
80 = 720/x
solve for x to get:
x = 720/80 = 9.
we get the same answer using both methods so we're good.
our data table becomes:
when s1 = 9, f1 = 80
when s2 = 6, f2 = 120
since 9*80 = 6*120, then the inverse relationship is good.
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