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Chapter 2: Q62P (page 501)

A vertical, \(1.20\;{\rm{m}}\) length of 18 gauge (diameter of 1.024 mm) copper wire has a 100.0 N ball hanging from it.

(a) What is the wavelength of the third harmonic for this wire?

(b) A 500.0 N ball now replaces the original ball. What is the change in the wavelength of the third harmonic caused by replacing the light ball with the heavy one? (Hint: Se Table 11.1 for Young’s modulus.)

Short Answer

Expert verified

Thus, (a) the wavelength of the third harmonic for the wire is 0.8 m

Step by step solution

01

(a) Given in the question

Tension in wire \(F = 100\;{\rm{N}}\).

Length of string \(L = 1.20\;m\)

Diameter of wire \(d = 1.024\;{\rm{mm}}\).

02

Use formula of wavelength

The formula of wavelength is:

\(\lambda = \frac{{2L}}{n}\).

Here,\(L\)length of string and\(n\)is harmonic number.

03

Calculate the wavelength

Solve the wavelength as follows:

\(\begin{array}{c}\lambda = \frac{{2L}}{n}\\ = \frac{2}{3} \times 1.2\\ = 0.8\;m\end{array}\)

Thus, the wavelength of the third harmonic for the wire is 0.8 m.

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