An organ is in tune at $22.0 ^\circ \textrm{C}$. By what percent will the frequency be off at $11 ^\circ \textrm{C}$?

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Initially the fundamental frequency in the organ tube is the initial speed of sound at this particular temperature, 22 degrees Celsius, divided by 2 times the organ tube length. And then the final frequency will be the new speed of sound divided by 2 times the same length as the organ tube itself hasn't changed. And so the percent difference between these two frequencies will be the absolute difference subtracting the frequencies divided by the initial frequency times 100%. And you can divide it to the top terms by the bottom and you get the ratio of frequencies minus 1 times 100%. So, this is f1 final, fundamental at the final temperature, 331 plus 0.6 times final temperature over 2 l. And then dividing it by fundamental initial is the same as multiplying by the reciprocal of this. So, we'll multiply by 2 l over 331 plus 0.6 times t initial. And the 2 l's cancel, leaving us with this formula. And that's going to be 331 plus 0.6 times 11 degrees Celsius for the final temperature, divided by 331 plus 0.6 times 22 degrees Celsius, initial temperature, all minus 1 times 100% which gives about negative 1.9% difference at 11 degrees Celsius.