Question

A string vibrates at its third-harmonic frequency. The amplitude at a point 30 cm from one end is half the maximum amplitude. How long is the string?

Short answer

The length of the string is 5.40 m

Step-by-step solution

Write the given information 

Third-harmonic frequency, n= 3
The amplitude of the wave at x=30cm is A_max/2

To determine the length of the string

The general expression of the amplitude of the wave at any distance x
$A\left(x\right)=A_{max}\sin \left(\frac{2\pi x}{\lambda }\right)$

According to the question
$$\begin{gathered} \frac{A_{max}}{2}=A_{max}\sin \left(\frac{2\pi (30cm)}{\lambda }\right) \\ \sin \left(\frac{60\pi }{\lambda }\right)=\frac{1}{2} \\ \Rightarrow \frac{60\pi }{\lambda }=\frac{\pi }{6} \\ \Rightarrow \lambda =360cm \end{gathered}$$

Since it is three harmonic vibrations, there would be three loops with the length λ/2.
Therefore, the total length of the string is given by
$L=3\left(\frac{360}{2}\right)=540cm$
Thus the length of the string is 5.40 m