Answer:
Part a)
f = 1911.5 Hz
Part b)
[tex]\lambda = 0.186 m[/tex]
Explanation:
Here the source and observer both are moving towards each other
so we know that the apparent frequency is given as
[tex]f' = f_0 (\frac{v + v_o}{v - v_s})[/tex]
here we know that
[tex]f_0 = 1750 Hz[/tex]
[tex]v_o = 15 m/s[/tex]
[tex]v_s = 15 m/s[/tex]
now we will have
[tex]f' = (1750)(\frac{340 + 15}{340 - 15})[/tex]
[tex]f' = 1911.5 Hz[/tex]
Part b)
Apparent wavelength is given by the formula
[tex]\lambda = \frac{v_{relative}}{f_{app}}[/tex]
here we will have
[tex]\lambda = \frac{340 + 15}{1911.5}[/tex]
[tex]\lambda = 0.186 m[/tex]