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A Levels Physics (9702)•9702/11/O/N/23
Question 22 from 9702/11/O/N/23

Explanation

Doppler Effect for Moving Source Steps: - Use the Doppler formula for source moving away from stationary observer: f′=fvv+vsf' = f \frac{v}{v + v_s}f′=fv+vs​v​, where f=800f = 800f=800 Hz, f′=750f' = 750f′=750 Hz, v=340v = 340v=340 m/s. - Rearrange: vs=v(ff′−1)=340(800750−1)=340(1615−1)=340×115≈22.7v_s = v \left( \frac{f}{f'} - 1 \right) = 340 \left( \frac{800}{750} - 1 \right) = 340 \left( \frac{16}{15} - 1 \right) = 340 \times \frac{1}{15} \approx 22.7vs​=v(f′f​−1)=340(750800​−1)=340(1516​−1)=340×151​≈22.7 m/s. - Approximate using Δff≈−vsv\frac{\Delta f}{f} \approx -\frac{v_s}{v}fΔf​≈−vvs​​: vs≈−Δffv=50800×340=21.25v_s \approx -\frac{\Delta f}{f} v = \frac{50}{800} \times 340 = 21.25vs​≈−fΔf​v=80050​×340=21.25 m/s away. - Options approximate to 21 m/s away, as exact 22.7 is closest to 21 (yields ~754 Hz). Why B is correct: - Matches approximate Doppler shift for source receding, where frequency decreases proportionally to source speed away from observer. Why the others are wrong: - A: 19 m/s away yields ~758 Hz, too high for 750 Hz. - C: 21 m/s towards yields ~852 Hz, frequency increases, not decreases. - D: 19 m/s towards yields ~849 Hz, frequency …

Topic: Doppler effect for sound waves

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