Q62P
Question
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.)
Step-by-Step Solution
VerifiedThus, (a) the wavelength of the third harmonic for the wire is 0.8 m
Tension in wire \(F = 100\;{\rm{N}}\).
Length of string \(L = 1.20\;m\)
Diameter of wire \(d = 1.024\;{\rm{mm}}\).
The formula of wavelength is:
\(\lambda = \frac{{2L}}{n}\).
Here, \(L\) length of string and \(n\) is harmonic number.
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.