Doppler Shift

To determine the received sound frequency when a bat at rest sends ultrasonic waves (50.0 kHz) that reflect off an object moving away at 22.0 m/s, we apply the Doppler effect twice^[1]: first when the object receives the waves (as a moving observer) and second when it reflects the waves (as a moving source). The calculation proceeds as follows:

Key Steps:

1. First Doppler Shift (Object as Observer):
   The object moves away from the bat, reducing the observed frequency:

   $$f^{\prime }=f_0\cdot \frac{v_{\text{sound}}-v_{\text{object}}}{v_{\text{sound}}}$$

2. Second Doppler Shift (Object as Source):
   The object now acts as a source moving away from the bat, further reducing the reflected frequency:

   $$f^{″}=f^{\prime }\cdot \frac{v_{\text{sound}}}{v_{\text{sound}}+v_{\text{object}}}$$

3. Combined Formula:
   Substituting $f^{\prime }$ into $f^{″}$:

   $$f^{″}=f_0\cdot \frac{v_{\text{sound}}-v_{\text{object}}}{v_{\text{sound}}+v_{\text{object}}}$$

Calculation:

• Given:
  $f_0=50.0\text{kHz},v_{\text{object}}=22.0\text{m/s},v_{\text{sound}}=343\text{m/s}$
• Result:$$f^{″}=50.0\text{kHz}\cdot \frac{343-22}{343+22}=50.0\text{kHz}\cdot \frac{321}{365}\approx 43.97\text{kHz}$$

Final Answer:

Footnotes: