Answer
The child is swinging up to a height of 2.3 meters.
Work Step by Step
Let $v_c$ be the child's speed at the bottom of the swing. We can use the equation for the Doppler effect when the observer is approaching to find an expression for the higher frequency $f_h$:
$f_h = \left(\frac{v+v_c}{v}\right)~f_s$
We can use the equation for the Doppler effect when the observer is moving away to find an expression for the lower frequency $f_l$:
$f_l = \left(\frac{v-v_c}{v}\right)~f_s$
Note that $f_h = 1.04~f_l$:
$f_h = 1.04~f_l$
$\left(\frac{v+v_c}{v}\right)~f_s = 1.04\left(\frac{v-v_c}{v}\right)~f_s$
$v+v_c = 1.04~(v-v_c)$
$2.04~v_c = 0.04~v$
$v_c = \frac{0.04~v}{2.04}$
$v_c = \frac{(0.04)~(343~m/s)}{2.04}$
$v_c = 6.725~m/s$
The speed of the child at the bottom of the swing is $6.725~m/s$
The kinetic energy at the bottom of the spring is equal to the potential energy at the highest point. We can find the height $h$ that the child is swinging:
$mgh = \frac{1}{2}mv^2$
$h = \frac{v^2}{2g}$
$h = \frac{(6.725~m/s)^2}{(2)(9.80~m/s^2)}$
$h = 2.3~m$
The child is swinging up to a height of 2.3 meters.
