Question

A rope, under a tension of $200N$ and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by : $y=0.1\sin \left(\frac{\pi x}{2}\right)\sin 12\pi t$.
Where $x=0$ at one end of the rope, $x$ is in meters and $t$ is in seconds. If the rope oscillates in a third-harmonic standing wave pattern, the period of oscillation is $1/x$ sec. Find $x$.

Answer 1

In the second harmonic, $\omega =12\pi$.

Hence the time period $T_2=\frac{2\pi }{\omega }=\frac{1}{6}$

$⟹$ Frequency=$\frac{1}{T_2}=6={\nu }_2$

For a standing wave pattern, ${\nu }_n=\frac{n}{2l}\sqrt{\frac{T}{\mu }}\propto n$

$⟹{\nu }_3=\frac{3}{2}{\nu }_2=9$

$⟹T_3=\frac{1}{9}$