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Chapter 15: Problem 38

A string is \(35.0 \mathrm{~cm}\) long and has a mass per unit length of \(5.51 \cdot 10^{-4} \mathrm{~kg} / \mathrm{m}\). What tension must be applied to the string so that it vibrates at the fundamental frequency of \(660 \mathrm{Hz?}\)

Short Answer

Expert verified
Answer: The tension that must be applied to the string is approximately 56.38 N.

Step by step solution

01

Ensure all values are in the correct units

Since the length of the string, \(L\) is given in cm, we need to convert it into meters. The mass per unit length is given in kg/m, which is already in the correct units. \(L = 35.0 \mathrm{~cm} \times \frac{1 \mathrm{~m}}{100 \mathrm{~cm}} = 0.35 \mathrm{~m}\) Now we have all the values we need: Length of the string, L = 0.35 m Mass per unit length, μ = \(5.51 \times 10^{-4} \mathrm{~kg/m}\) Fundamental frequency, \(f_1 = 660 \mathrm{~Hz}\)
02

Solve for tension T using the formula

We're solving for \(T\) using the formula: $$f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$$ First, we'll isolate T by squaring both sides of the equation and then multiply by the denominator: $$T = \mu(2Lf_1)^2$$ Next, we'll plug in the values we have for L, μ, and \(f_1\): $$T = (5.51 \times 10^{-4} \mathrm{~kg/m})\left(2\times(0.35 \mathrm{~m})\times(660 \mathrm{~Hz})\right)^2$$
03

Calculate the tension T

Now, we can perform the calculations: $$T = (5.51 \times 10^{-4} \mathrm{~kg/m}) \times (4\times0.35 \mathrm{~m}\times660 \mathrm{~Hz})^2$$ $$T ≈ 56.38 \mathrm{~N}$$ This gives the tension in the string that must be applied for it to vibrate at the fundamental frequency of 660 Hz.

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