An instrument consists in a series of long metal cylinders, each of length x and

Question

An instrument consists in a series of long metal cylinders, each of length x and
diameter x/10. The cylinders are closed at one end and open at the other. A string
is stretched across the open end. The string has mass per unit length µ and tension
T. The string is put under enough tension so that the fundamental frequency of the
waves in the vibrating string equals the n=3 mode of the sound waves in the
cylinder. Derive a formula for the tension in the string in terms of the other
parameters of the instrument and the speed of sound in air, v.

Explanation / Answer

there are two things need to be consider, the string and the cylinder. Consider the string over the open-closed air column. Since the string is going across the diameter of the cylinder, the length of the string will be the diameter of the column. L_string = x/10 Next we look at the equation that I'm sure you know, v_string = f_string * lambda where we can rearrange the equation to f = 3v/lambda, (the 3 came from the fact that it is an open-closed air column with the 3rd harmonic, based on the equation f = mv/lambda where m = 1,3,5...) We know that the wavelength of the string will be vibrating at the resonance frequency so lambda = 2L = 2(x/10) = x/5 Sub in the equation v = sqrt(T/(mu)) (known from the previous unit) to the first derived equation and we now have: f = 15sqrt(T/(mu))/x Now we know that the frequency from the string is the same as the frequency from the wavelength, so now we just have to compare frequencies. The frequency of the cylinder is going to be f = v/4x (since the column is open-closed, so we know that lambda will be equal to 4L, where L is going to be x) Now just set the first derived equation equal to the second one, and solve for T. 15sqrt(T/(mu))/x = v/4x 15sqrt(T/(mu)) = v/4 sqrt(T/(mu)) = v/60 T =1/3600( v^2*(mu) )

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