One end of a string of length 'L' is tied to ceiling of a lift accelerating upwards with an acceleration ' 2g'. The other end of the string is free. The linear mass density of the string varies linearly from '0' to '$\mathrm{\lambda }$' from bottom to top. A transverse pulse is generated near the bottom of string.

$\mathrm{V}=\frac{\mathrm{dx}}{\mathrm{dt}}=\sqrt{\frac{\mathrm{T}}{\mathrm{\mu }}}$

Here $\mathrm{\mu }=\frac{\mathrm{\lambda }}{\mathrm{L}}x$

where x is distance from bottom.

Tension in the string at some point x from the bottom is 

$\mathrm{T}={\mathrm{M}}_{\mathrm{x}}\mathrm{a}=3{\mathrm{M}}_{\mathrm{x}}\mathrm{g} (\because \mathrm{lift} \mathrm{is} \mathrm{moving} \mathrm{up}, \mathrm{a} = 2\mathrm{g})$

${\mathrm{M}}_{\mathrm{x}}= {\int }_0^{\mathrm{x}}\frac{\mathrm{\lambda }}{\mathrm{L}}xdx$

${\mathrm{M}}_{\mathrm{x}}=\frac{{\mathrm{\lambda x}}^2}{2\mathrm{L}}$

$\mathrm{T}=2{\mathrm{M}}_{\mathrm{x}}\mathrm{g}=\frac{3{\mathrm{\lambda x}}^2}{2\mathrm{L}}\mathrm{g}$

$\Rightarrow \mathrm{V}=\sqrt{\frac{\mathrm{T}}{\mathrm{\mu }}}=\sqrt{\frac{3\mathrm{xg}}{2}}$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\sqrt{\frac{3\mathrm{xg}}{2}}\Rightarrow \frac{\mathrm{dx}}{\sqrt{\mathrm{x}}}=\sqrt{\frac{3\mathrm{g}}{2}}\mathrm{dt}$

$\Rightarrow 2\sqrt{\mathrm{x}}=\sqrt{\frac{3\mathrm{g}}{2}}\mathrm{t}$..............(1)

From this equation, time taken for the pulse to reach the top is 

$\mathrm{T}=\sqrt{8\mathrm{L}/3\mathrm{g}}$

Also from (1) :

$x=\frac{3g}{8}{t}^2$

$\Rightarrow \mathrm{a}=\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}=\frac{3\mathrm{g}}{4}$relative to the string.