${y_1} = {10^{ - 6}}\sin [100\,t + (x/50) + 0.5]m$
${y_2} = {10^{ - 6}}\cos \,[100\,t + (x/50)]m$
where $ x$ is expressed in metres and $t$ is expressed in seconds, is approximately .... $ rad$
${y_2} = {10^{ - 6}}\sin \,\left[ {100\,t + \left( {\frac{x}{{50}}} \right) + \left( {\frac{\pi }{2}} \right)} \right]$
Phase difference $\phi$
$=[100 t + (x/50) + 1.57]-[100 t + (x/50)+ 0.5]$
$ = 1.07\,radians.$
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($G$ is the gravitational constant: $\mathrm{M}$ is the mass of the earth)