# Question #d5420

May 2, 2017

Because the ends are fixed, standing waves can be made with any integer number of half wavelengths, ie for rope length $L$:

$L = n {\lambda}_{n} / 2 , n \in {\mathbb{Z}}^{+} \implies {\lambda}_{n} = \frac{2 L}{n}$.

Comparing the 5m and 4m wavelengths, which we are told are consecutive harmonics, ie they are ${\lambda}_{n}$ and ${\lambda}_{n + 1}$:

$\frac{5}{4} = \frac{\frac{2 L}{n}}{\frac{2 L}{n + 1}} \implies n = 4$

$\implies {\lambda}_{4} = 5 = \frac{2 L}{4} \implies L = 10 \text{ m}$

And to check:

$\implies {\lambda}_{5} = 4 = \frac{2 L}{4 + 1} \implies L = 10 \text{ m}$

For (ii), $M = \rho L = 6 \text{ kg}$.