Dawson 203-SN3-RE tutoring

1. Interpret the mode. A node is a point whose transverse displacement stays zero; an antinode is a point of maximum oscillation amplitude. Each fixed end must be a node. Let $n$ denote the positive-integer harmonic number and $\lambda$ the wavelength. A string fixed at both ends contains $n$ half-wavelengths, so $L=n\lambda/2$. For the third harmonic, $n=3$, giving $\lambda=2L/3=0.50\,\mathrm{m}$. 2. Find frequency. Let $f$ denote the number of oscillations per second. The wave relation $v=f\lambda$ gives $f=v/\lambda=(90\,\mathrm{m/s})/(0.50\,\mathrm{m})=180\,\mathrm{Hz}$. Three half-wavelengths, rather than three full wavelengths, fit between the fixed ends. 3. Locate nodes and antinodes. Adjacent nodes are separated by $\lambda/2=0.25\,\mathrm{m}$, so the node positions are $x=0$, $0.25$, $0.50$, and $0.75\,\mathrm{m}$. Each antinode lies halfway between adjacent nodes: $x=0.125$, $0.375$, and $0.625\,\mathrm{m}$. These positions give three vibrating segments and include both endpoints among the nodes, as required. 4. Change the tension. Let $T$ denote the string tension in newtons and $\mu$ its linear mass density, meaning mass per unit length in $\mathrm{kg/m}$. For this ideal string, $v=\sqrt{T/\mu}$. Replacing $T$ by $4T$ with $\mu$ fixed multiplies speed by $\sqrt4=2$, giving $180\,\mathrm{m/s}$. Keeping $L$ and $n=3$ fixed preserves $\lambda=2L/3$, so the node and antinode positions stay unchanged. Frequency doubles to $360\,\mathrm{Hz}$ because $f=v/\lambda$. This changes the resonant frequency; it does not determine the vibration amplitude.