## Essential University Physics: Volume 1 (3rd Edition)

$E =\frac{4\pi^2FA^2}{\lambda}$
We know that the energy is given by: $E=\mu \omega^2 A^2 vT$ $E=\frac{F}{v^2} \omega^2 A^2 vT$ $E=\frac{F}{v} \omega^2 A^2 T$ $E=\frac{F}{v} (\frac{2\pi}{T})^2 A^2 T$ $E=\frac{F}{v} (\frac{4\pi^2}{T}) A^2$ $E =\frac{4\pi^2FA^2}{\lambda}$