Question

a guitar string is 90 cm long and has a mass of 3.8 g. the distance from the bridge to the support post is ( l = 62 ) cm, and the string is under a tension of 540 n. part a what are the frequencies of the fundamental and first two overtones? express your answers using two significant figures. enter your answers in ascending order separated by commas. ( f_1,f_2,f_3= ) hz

Explanation:

Step1: Calculate the linear mass density

The linear mass density \(\mu=\frac{m}{L_{string}}\). Given \(m = 3.8\ g=3.8\times10^{- 3}\ kg\) and \(L_{string}=90\ cm = 0.9\ m\), so \(\mu=\frac{3.8\times 10^{-3}}{0.9}\ kg/m\approx4.22\times10^{-3}\ kg/m\)

Step2: Calculate the wave speed

The wave speed \(v=\sqrt{\frac{T}{\mu}}\), where \(T = 540\ N\) and \(\mu\approx4.22\times10^{-3}\ kg/m\). Then \(v=\sqrt{\frac{540}{4.22\times10^{-3}}}\ m/s\approx358\ m/s\)

Step3: Calculate the fundamental frequency

For a vibrating string fixed at both ends, the fundamental frequency \(f_1=\frac{v}{2L}\), with \(L = 62\ cm=0.62\ m\) and \(v\approx358\ m/s\). So \(f_1=\frac{358}{2\times0.62}\ Hz\approx290\ Hz\)

Step4: Calculate the first overtone

The first overtone \(f_2 = 2f_1\). Since \(f_1\approx290\ Hz\), then \(f_2=2\times290\ Hz = 580\ Hz\)

Step5: Calculate the second overtone

The second overtone \(f_3=3f_1\). Since \(f_1\approx290\ Hz\), then \(f_3 = 3\times290\ Hz=870\ Hz\)

Answer:

\(290,580,870\)