Twang!

The speed of the wave increases as it travels up the string, because higher parts of the string carry more weight and therefore have more tension. We integrate $dt$ over the length of the string, starting with $x = 0$ at the bottom.

Note that the wave speed in a string is $v = \sqrt{F/\mu}$ , where $\mu = m/\ell$ and $F$ is the weight hanging below this point of the string. I find it convenient to define a "length" $L = \frac M \mu = \frac M m \ell.$ (This can be interpreted as the length of string that would have the same mass as the hanging mass.)

$t = \int_0^\ell \frac{dx}v = \int_0^\ell dx \sqrt{\frac \mu F} \\ = \int_0^\ell dx \sqrt{\frac \mu {(M + \mu x)g}} \\ = \frac1{\sqrt g} \int_0^\ell dx\sqrt{\frac 1{L + x}} \\ = \frac 2{\sqrt g} \left.\sqrt{L + x}\right|_0^\ell \\ = \frac 2{\sqrt g}\left(\sqrt{L + \ell} - \sqrt L\right) \\ = \frac 2{\sqrt g}\left(\sqrt{\frac{(M + m)\ell} m} - \sqrt {\frac{M\ell} m}\right) \\ = \color{#D61F06}{2\frac{\sqrt{M+m}-\sqrt{M}} {\sqrt m}} \sqrt{\frac{\ell}{g}}.$

Note that the answer $c = \sqrt{m/M}$ would be correct if the tension in the string were a constant $F = Mg$ .