A police siren emits a sinusoidal wave with frequency `f_(S) = 300 H_(z)` . The speed of the sound is `340 m//s`. (a) Find the wavelength of the waves if siren is at rest in the air . (b) If the siren is moving at `30 m//s`, Find the wavelength of the waves ahead of and behind the source.

Explanation

The doppler effect is not involved in part (a), since neither the source nor the listener is moving. In part (b), the source is in motion and we must involve the doppler effect. Figure. Shows the situation .We use the relation ship `v=lamdaf`
to determine the wavelength when the polive siren is at rest When it is in motion, we find the wavelength on either side of siren.
a. When the source is at rest,
`lamda=(v)/(f_S)=(340(m)/(s))/(300Hz)=1.13m`
b. The situation is shown in Fig. Now, in front of the siren,
`lamda_(in front)=(v-v_S)/(f_S)=(340(m)/(s)-30(m)/(s))/(300Hz)=1.03m`
Behind the siren,
`lamda_(behind)=(v+v_S)/(f_S)=(340(m)/(s)+30(m)/(s))/(300Hz)=1.23m`
The wavelength is less in front of the siren and greater behind the siren, as it should be.