Answer:
The answer is about 0,59 m or 59 cm the longest wavelength for a listener when the sirens is closest.
Explanation:
Using the equation of the effect Doopler:
[tex]f'= \frac{(V + V_{L}) }{(V + V_{s})}*f_{o}[/tex]
Replacing:
W angular velocity = 0,8 [tex]\frac{rad}{s}[/tex]
V = 350 [tex]\frac{m}{s}[/tex]
[tex]f_{o} = 600 Hz[/tex]
VL = 0 [tex]\frac{m}{s}[/tex] [tex] ( Is the velocity of the listener, we assume he is at rest)
Vs = W*Radius ⇒ Vs = 0,8 [tex]\frac{rad}{s}[/tex] * 5 m
Vs = 4 [tex]\frac{m}{s}[/tex]
Replacing:
[tex]f'= \frac{(V + V_{L}) }{(V + V_{s})}*f_{o}[/tex]
[tex]f'= \frac{350 \frac{m}{s} }{(350 + 4) \frac{m}{s} } * 600 Hz[/tex]
Note: Notice, the division have the same units so, it can be simplify
[tex]f'= 593,220339 Hz[/tex]
Now using the frequency f' as the frequency of the listener to know L wavelength as the relation between f' and V
[tex]L = \frac{V}{f'}[/tex]
[tex]L = \frac{350 \frac{m}{s} }{ 593,220339 Hz}[/tex]
Note: Remember Hz units are relations inverse of time so: [tex]Hz = \frac{1}{s}[/tex]
[tex]L = \frac{350 \frac{m}{s} }{ 593,220339 \frac{1}{s}}[/tex]
[tex]L = 0,59 m, or, 0,59 cm[/tex]
