Chapter 15.1, Problem 5PP

To determine

To calculate: The velocity of the trumpet to raise the pitch to $277\text{ Hz}$ .

Answer to Problem 5PP

The trumpet would have to move at $18.60\text{ m/s}$ to raise the pitch to $277\text{ Hz}$ .

Explanation of Solution

Given information:

Speed of sound in air ( $v$ ) = $343\text{ m/s}$

Frequency of wave ( $f_s$ ) = $262\text{ Hz}$

Frequency perceived by a detector ( $f_d$ ) = $277\text{ Hz}$

Formula used:

Doppler Effect equation:

  $f_d=f_s\left(\frac{v-v_d}{v-v_s}\right)$

Where $f_d$ represents frequency perceived by the detector, $f_s$ represents wave’s frequency, $v$ represents speed of sound in air, $v_s$ represents velocity of source and $v_d$ represents velocity of detector.

Calculations:

As per problem,

Assuming, the detector is at rest, $v_d=0\text{ m/s}$ and the source is moving towards the detector.

To calculate the speed of the trumpet (source), rearrange above stated formula

  $\begin{array}{l}{f}_d=f_s\left(\frac{v-v_d}{v-v_s}\right)\\ v_s=v-\frac{(v-v_d)f_s}{f_d}\end{array}$

Substitute the values

  $\begin{array}{l}v=343\text{ m/s}\\ v_d=0\text{ m/s}\\ f_s=262\text{ Hz}\\ f_d=277\text{ Hz}\end{array}$

  $\begin{array}{l}{v}_s=v-\frac{(v-v_d)f_s}{f_d}\\ v_s=343\text{ m/s}-\frac{(343\text{ m/s}-0\text{ m/s})(262\text{ Hz)}}{277\text{ Hz}}\\ v_s=343\text{ m/s}-\frac{(343\text{ m/s)(262 Hz)}}{277\text{ Hz}}\\ v_s=343\text{ m/s}-321.9\text{ m/s}\\ v_s=18.6\text{ m/s}\end{array}$

Conclusion:

Thus, the trumpet would have to move at $18.60\text{ m/s}$ to raise the pitch to $277\text{ Hz}$ .