Question

a string has a mass of 12.3 g. the string is stretched with a force of 8.63 n, giving it a length of 1.91 m. then, the string vibrates transversely at precisely the frequency that corresponds to its fourth normal mode; that is, at its fourth harmonic.
what is the wavelength λ of the standing wave created in the string?
what is the frequency f of the standing wave?

Explanation:

Step1: Find the wavelength

For a string fixed at both ends, the formula for the wavelength in the \(n^{th}\) harmonic is \(\lambda=\frac{2L}{n}\). Here, \(n = 4\) (fourth harmonic) and \(L=1.91\ m\).

$$ \lambda=\frac{2\times1.91}{4} $$

$$ \lambda = 0.955\ m $$

Step2: Find the linear mass density

The linear mass density \(\mu=\frac{m}{L}\), where \(m = 12.3\ g=0.0123\ kg\) and \(L = 1.91\ m\).

$$ \mu=\frac{0.0123}{1.91}\ kg/m\approx0.00644\ kg/m $$

Step3: Find the wave speed

The wave speed \(v=\sqrt{\frac{F}{\mu}}\), where \(F = 8.63\ N\) and \(\mu\approx0.00644\ kg/m\).

$$ v=\sqrt{\frac{8.63}{0.00644}}\approx36.7\ m/s $$

Step4: Find the frequency

Using the formula \(v = f\lambda\), we can solve for \(f\). Since \(v\approx36.7\ m/s\) and \(\lambda = 0.955\ m\)

$$ f=\frac{v}{\lambda}=\frac{36.7}{0.955}\approx38.4\ Hz $$

Answer:

\(\lambda = 0.955\ m\)
\(f\approx38.4\ Hz\)