Answer
$v_{heart}=0.082m/s$
Work Step by Step
Use equation $f'=f\Big(\frac{v_{snd}\pm v_{obs}}{v_{snd}\mp v_{source}}\Big)$, upper signs if moving toward each other and lower signs if moving away from each other.
1. Heart is observer moving away from or toward sound source. $v_{source}=0m/s$
$f'=f\Big(\frac{v_{snd}+ v_{obs}}{v_{snd}}\Big)$ for heart moving toward source
2. Heart is the sound source moving toward or away from observer at rest $v_{obs}=0m/s$
$f''=f'\Big(\frac{v_{snd}}{v_{snd}-v_{source}}\Big)$ for heart moving toward observer
$f''=f\Big(\frac{v_{snd}+ v_{obs}}{v_{snd}}\Big)\Big(\frac{v_{snd}}{v_{snd}-v_{source}}\Big)$
$f''=f\Big(\frac{v_{snd}+ v_{obs}}{v_{snd}-v_{source}}\Big)$
Beat frequency $\Delta f = f''-f=f\Big(\frac{v_{snd}+ v_{obs}}{v_{snd}-v_{source}}\Big)-f$
$v_{heart} = v_{obs}=v_{source}$
$\Delta f =f\Big(\frac{v_{snd}+ v_{heart}}{v_{snd}-v_{heart}}\Big)-f$
$240Hz=(2.25\times10^6Hz)\Big(\frac{1540m/s+ v_{heart}}{1540m/s-v_{heart}}-1\Big)$
$v_{heart}=0.082m/s$
Regardless of whether heart is taken to be moving away from or toward the observer and source, the difference in the answer is within the range of the number of significant figures and can be ignored.
