Solution:
Solution:
Observer is stationary and source is moving.
During approach, $$f_1=f\frac{v}{v-v_s}$$
$$\hspace30mm = 1000 \bigg(\frac{320}{320-20}\bigg)=1066.67\, Hz$$
During recede, $$f_2=f\bigg(\frac{v}{v+v_s}\bigg)$$
$$\hspace30mm = 1000 \bigg(\frac{320}{320+20}\bigg)=941.18\, Hz$$
|$$\%$$ change in frequency| $$= \bigg(\frac{f_1-f_2}{f_1}\bigg) \times 100 $$
$$\hspace25mm = 12 \%$$