DOPPLER EFFECT I: WAVELENGTHS
A police car’s siren emits a sinusoidal wave with frequency f_S = 300 Hz. The speed of sound is 340 m/s and the air is still. (a) Find the wavelength of the waves if the siren is at rest. (b) Find the wavelengths of the waves in front of and behind the siren if it is moving at 30 m/s.
IDENTIFY and SET UP:
In part (a) there is no Doppler effect because neither source nor listener is moving with respect to the air; v = λf gives the wavelength. Figure 16.30 shows the situation in part (b): The source is in motion, so we find the wavelengths using Eqs. (16.27) and (16.28) for the Doppler effect.
\lambda_{\text {in front }}=\frac{v}{f_{\mathrm{S}}}-\frac{v_{\mathrm{S}}}{f_{\mathrm{S}}}=\frac{v-v_{\mathrm{S}}}{f_{\mathrm{S}}} (16.27)
\lambda_{\text {behind }}=\frac{v+v_{\mathrm{S}}}{f_{\mathrm{S}}} (16.28)
EXECUTE:
(a) When the source is at rest,
\lambda=\frac{v}{f_{\mathrm{S}}}=\frac{340 \mathrm{~m} / \mathrm{s}}{300 \mathrm{~Hz}}=1.13 \mathrm{~m}(b) From Eq. (16.27), in front of the siren
\lambda_{\text {in front }}=\frac{v-v_{\mathrm{S}}}{f_{\mathrm{S}}}=\frac{340 \mathrm{~m} / \mathrm{s}-30 \mathrm{~m} / \mathrm{s}}{300 \mathrm{~Hz}}=1.03 \mathrm{~m}From Eq. (16.28), behind the siren
\lambda_{\text {behind }}=\frac{v+v_{\mathrm{S}}}{f_{\mathrm{S}}}=\frac{340 \mathrm{~m} / \mathrm{s}+30 \mathrm{~m} / \mathrm{s}}{300 \mathrm{~Hz}}=1.23 \mathrm{~m}
EVALUATE: The wavelength is shorter in front of the siren and longer behind it, as we expect.
