Answer :
Answer:
304.34 m/s
Explanation:
We are given that
Frequency of middle C=f=261.7 Hz
Frequency of B4 emit,f'=493.9 Hz
We have to find the speed of harmonica must have in order for a stationary observer to hear 493.9 Hz.
Speed of sound,v=343 m/s
We know that
[tex]f'=f(\frac{v+v'}{v})[/tex]
[tex]493.9=261.7(\frac{343+v'}{343})[/tex]
[tex]343+v'=\frac{493.9\times 343}{261.7}[/tex]
[tex]v'=\frac{493.9\times 343}{261.7}-343[/tex]
[tex]v'=304.34m/s[/tex]