Physics - Wave on String Question with Solution | TestHub

Comparing given equation with standard equation $y=2a\sin \left(\frac{2\pi x}{\lambda }\right)\cos (2\pi ft)$ where, $a$ is the amplitude, $k=\frac{2\pi }{\lambda }$ the propagation constant and $f$ the frequency. Therefore, we obtain: $\frac{2\pi }{\lambda }=\frac{\pi }{15}{\mathrm{cm}}^{-1}$ $\Rightarrow \lambda =30\mathrm{cm}$ Distance between nearest node and antinode $=\frac{\lambda }{4}=\frac{30}{4}=7.5\mathrm{cm}$