- #1
Spookie71
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Homework Statement
A guitar string of length 60.0 cm under 81.0 N of tension produces a note of frequency 330.0 Hz. What frequency will the same string produce when under 100.0 N of tension and shortened to 25.0 cm?
Homework Equations
Here is the question
[tex]f_{f}[/tex] = 330 Hz * [tex]\sqrt{ \frac{100.0 N}{81.0 N}}[/tex] * [tex]\frac{30.0 cm}{25.0 cm}[/tex]
The answer is 417.42 Hz in the book, I just don't know the order of operations to solve this.
The Attempt at a Solution
Can you explain where I'm going wrong in my order of operations.When I try it I take 100.0 N and divide by 81.0 N where I get 1.234567901 and then I push the square root key on my calculator which gives me 1.111111111
I then take 330 * by 1.111111111 * 1.2
Which gives me 440 Hz.
Thanks