Question

question 9 - An eagle flying at 28 m/s emits a cry whose frequency is 370 Hz. A blackbird is moving in the same direction as the eagle at 11.35 m/s. (Assume the speed of sound is 343 m/s). What frequency does the blackbird hear (in Hz) as the eagle approaches the blackbird?For the scenario in Question 9 with the speed of eagle 11.46 m/s, what frequency does the blackbird hear (in Hz) after the eagle passes the blackbird?

Answer 1

Given:

Velocity of eagle = 28 m/s

Frequency of eagle = 370 Hz

Velocity of black bird = 11.35 m/s

Speed of sound = 343 m/s

Let's find the frequency the black bird hears as the eagle approaches the black bird.

Apply the formula;

`f=f_e((v-v_b)/(v-v_e))`

Where:

fe is frequency of eagle = 370 Hz

v = 343 m/s

vb is the frequency of the blackbird= 11.35 m/s

v = 343 m/s

ve is the speed of eagle = 28 m/s

Substitute values into the formula and solve for f:

`\begin{gathered} f=370((343-11.35)/(343-28)) \\ \\ f=370((331.65)/(25)) \\ \\ f=389.56\text{ Hz} \end{gathered}`

Therefore, the frequency the blackbird hear as the eagle appoaches is 389.56 Hz.

(b). Given:

Speed of eagle = 11.46.

What frequency does the blackbird hear (in Hz) after the eagle passes the blackbird?

Apply the formula:

`f=f_b((v+v_b)/(v+v_e))`

Where:

ve = 11.46 m/s

Thus, we have:

`\begin{gathered} f=370((343-11.35)/(343-11.46)) \\ \\ f=370.12\text{ Hz} \end{gathered}`

The frequency after the eagle passes the blackbird at this new speed is 370.12 Hz.

ANSWER:

• (a). 389.56 Hz.

• (b). 370.12 Hz.