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Chapter 11: Problem 53

A 1.6 -m-long string fixed at both ends vibrates at resonant frequencies of \(780 \mathrm{Hz}\) and \(1040 \mathrm{Hz}\), with no other resonant frequency between these values. (a) What is the fundamental frequency of this string? (b) When the tension in the string is \(1200 \mathrm{N},\) what is the total mass of the string?

Short Answer

Expert verified
The fundamental frequency of the string is approximately 346.67 Hz. (b) What is the total mass of the string when the tension is 1200 N? The total mass of the string is approximately 0.00326 kg (or 3.26 grams).

Step by step solution

01

Find the relationship between the resonant frequencies and the fundamental frequency

The resonant frequencies of a string fixed at both ends are given by $$f_n = (2n - 1)\frac{f_1}{2},$$ where \(f_n\) is the nth resonant frequency and \(f_1\) is the fundamental frequency. We know the subsequent resonant frequencies are 780 Hz and 1040 Hz, so we can write the given frequencies as: $$f_1 = 780 \mathrm{Hz}$$ $$f_2 = 1040 \mathrm{Hz}.$$
02

Find the fundamental frequency

Using the relationship found in step 1, we can write the equation for the fundamental frequency as: $$f_1 = \frac{780 \mathrm{Hz}}{1}$$ and $$f_1 = \frac{1040 \mathrm{Hz}}{3}.$$ Solving for the fundamental frequency, we get $$f_1 = \frac{1040 \mathrm{Hz}}{3} = 346.67 \mathrm{Hz}$$. So, the fundamental frequency of this string is approximately 346.67 Hz.
03

Use the wave equation to relate the fundamental frequency to the properties of the string

The wave equation for a string fixed at both ends is $$f = \frac{v}{2L},$$ where \(f\) is the frequency, \(v\) is the wave speed on the string, and \(L\) is the length of the string. The wave speed can also be expressed in terms of tension (\(T\)) and linear mass density (\(\mu\)) as: $$v = \sqrt{\frac{T}{\mu}}.$$
04

Find the linear mass density

We have found the fundamental frequency and the given tension in the string, so we can now find the linear mass density. Combine the two equations from Steps 3 and substitute in the values for \(f\), \(T\), and \(L\): $$f_1 = \frac{\sqrt{\frac{T}{\mu}}}{2L}$$ $$346.67 \mathrm{Hz} = \frac{\sqrt{\frac{1200 \mathrm{N}}{\mu}}}{2(1.6\mathrm{m})}$$ Solve for \(\mu\): $$\mu = \frac{1200 \mathrm{N}}{((2(1.6\mathrm{m})(346.67\mathrm{Hz}))^2} \approx 2.04 \times 10^{-3} \mathrm{kg/m}.$$ The linear mass density of the string is approximately \(2.04 \times 10^{-3}\) kg/m.
05

Find the total mass of the string

Now that we have the linear mass density, we can calculate the total mass of the string using the following equation: $$M = \mu L,$$ where \(M\) is the total mass of the string and \(L\) is the length of the string. Plug in the values for \(\mu\) and \(L\): $$M = (2.04 \times 10^{-3} \mathrm{kg/m})(1.6\mathrm{m}) \approx 0.00326 \mathrm{kg}.$$ The total mass of the string is approximately \(0.00326\) kg (or 3.26 grams).

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