Temperature and Frequency: Calculating Percent Difference for Piano Tuning

[Anony-mouse]

Homework Statement

A piano is to be tuned at 20.0°C. However, when the temperature has lowered to 6.32°C, the piano will be off tune. By what percentage will the piano be off key at this low temperature?

Homework Equations

V = Square Root (3RT/M)

The Attempt at a Solution

T must be in Kelvins, so 20--> 293 K and 6.32--> 279.47 K

Because 3R/M is constant for both temperatures, you can ignore that in the equation so it is just the square root of temperature.
V1 = Sqrt (293)
V2 = Sqrt (279.47)

To find the percent difference, I did (1 - V2/V1) = 0.0234. Thus, the percent difference is 2.34%, but that wasn't the right answer.

 

[Anony-mouse]

I just tried doing it a different way, but that still didn't give me the right answer.

V = 331 + .6(T)
V1=343 m/s
V2=334.792 m/s

1-v2/v1 = 0.0239 = 2.39%

I am really confused on this problem if anyone knows what I am doing wrong.

 

[mgb_phys]

Where did you get the equation from ?
Are you sure T isn't tension?

 

[ital_dj]

I'm no expert, I'm in grade 11 just like you but I'm a little better at this stuff than the rest... I'm not sure if I'm right, but use the V = 331 + .6(T) formula, and divide the first number (original), by the new number? And you might get a percentage?

Where did you get the equation from ?
Are you sure T isn't tension?

T is the temperature, it is also tension in different equations.
The equation that I repeated (v=) is from the Grade 11 textbook... it is a valid formula.

 

[Anony-mouse]

I flipped the equation so that it was V1/V2, which gave a percent of 2.45, but that didn't work either. I don't think this question is supposed to be that difficult, so could it just be a problem with the answer, since I'm pretty sure everything I did was right.

Does anyone notice anything wrong though with the work I did earlier?