## College Physics (4th Edition)

$v = \sqrt{g~\lambda}$
The wave speed $v$ is measured in units of $m~s^{-1}$ $g$ is measured in units of $m~s^{-2}$ $\lambda$ is measured in units of $m$ Let's assume that $v = g^a~\lambda^b$ Then: $m~s^{-1} = (m~s^{-2})^a~(m)^b$ We can consider the units of $s$: $(s^{-2})^a = s^{-1}$ $s^{-2a} = s^{-1}$ $-2a = -1$ $a = \frac{1}{2}$ We can consider the units of $m$: $(m)^a~(m)^b = m^1$ $a+b = 1$ $b = 1-a$ $b = 1-\frac{1}{2}$ $b = \frac{1}{2}$ We can use the exponents to write the equation: $v = g^a~\lambda^b$ $v = g^{1/2}~\lambda^{1/2}$ $v = \sqrt{g~\lambda}$