String Waves and Matter Waves — Question 2

Question 2

Interior nodes determine the mode index without counting the walls

A standing mode between two fixed endpoints has four interior nodes. The endpoint separation is $L=.750\,\mathrm m$. The string wave speed is $v=120\,\mathrm{m/s}$.

1. Identify the mode and its wavelength.

2. Find its frequency and the frequency difference to the next mode.

Question 2 - Solution

Interior nodes determine the mode index without counting the walls

(a) Method and calculation

$$N_{\mathrm{interior}}=n-1=4\Rightarrow\boxed{n=5}.$$$$\boxed{\lambda_5=2L/5=.300\,\mathrm m}.$$

(b) Interpretation and further calculation

$$\boxed{f_5=v/\lambda_5=400\,\mathrm{Hz}}.$$$$\boxed{f_6-f_5=v/(2L)=80.0\,\mathrm{Hz}}.$$

Physical check

The two boundary nodes are not included in the count of interior nodes. This equally spaced frequency ladder is specific to the nondispersive string.