How do I calculate the frequencies of the following?

[riseofphoenix]

7. A steel wire in a piano has a length of 0.9000 m and a mass of 4.000 10-3 kg. To what tension must this wire be stretched so that the fundamental vibration corresponds to middle C (fC = 261.6 Hz on the chromatic musical scale)?


8. The human ear canal is about 2.6 cm long. If it is regarded as a tube open at one end and closed at the eardrum, what is the fundamental frequency around which we would expect hearing to be most sensitive?

 

[riseofphoenix]

Do I use:

f = TM/L√2L

 

[riseofphoenix]

For the first one I got T = 49888794.74

 

[riseofphoenix]

Oh wait the first one is T = 985 N

 

[Nickie]

To calculate the frequency of the steel wire in the piano, we can use the equation f = (1/2L)√(T/μ), where f is the frequency, L is the length of the wire, T is the tension, and μ is the linear density (mass per unit length) of the wire. In this case, we are given the length (0.9000 m) and the mass (4.000 x 10^-3 kg) of the wire. To find the linear density, we can divide the mass by the length, giving us a value of 0.0044 kg/m. We also know that the desired frequency is 261.6 Hz. Rearranging the equation, we get T = (4μf^2L^2), where T is the tension we need to calculate. Plugging in the values, we get T = (4 x 0.0044 x 261.6^2 x 0.9000^2) = 262.6 N. Therefore, the tension required for the steel wire to vibrate at middle C is 262.6 Newtons.

To calculate the fundamental frequency of the human ear canal, we can use the equation f = v/4L, where f is the frequency, v is the speed of sound (343 m/s in air), and L is the length of the ear canal. Plugging in the values, we get f = 343/(4 x 0.026) = 3290.4 Hz. However, it's important to note that this is just an estimation and the actual fundamental frequency may vary depending on the individual's ear canal shape and size. Additionally, the human ear is sensitive to a range of frequencies, not just the fundamental frequency.