Chapter 22 - Vibrations - Section 22.6 - Electrical Circuit Analogs - Problems - Page 676: 60

$1.2m/s$

We can find the required speed as follows: $\omega_n=\sqrt{\frac{K}{m}}$ $\implies \omega_n=\sqrt{\frac{2\times 800}{450}}$ $\implies \omega_n=1.8856rad/s$ We know that $T=\frac{2\pi}{\omega_n}$ $\implies T=\frac{2\pi}{1.8856}$ $\implies T=3.3322s$ Now $v=\frac{\lambda}{T}$ We plug in the known values to obtain: $v=\frac{4}{3.3322}$ $\implies v=1.2m/s$