Question

The sound of a racing car has its pitch decreased by 10%. If the temperature of the air is 22oC, how fast is the car travelling?

Answer 1

The speed of sound at 22 degrees Celsius is v = 344.31 m/s.

The pitch, p, and frequency, f are related as

`\text{p }\propto f`

If pitch increases, then frequency increases, and if pitch decreases the frequency decreases.

The pitch decreased by 10%, so the frequency also decreases by 10%.

Let the original frequency be f.

The observed frequency will be

`\begin{gathered} f^(\prime)=f-(10f)/(100) \\ =\text{ }(9f)/(10) \end{gathered}`

Let the speed of the car be v' which can be calculated by the formula

`\begin{gathered} f^(\prime)=((v)/(v+v^(\prime)))f \\ v^(\prime)=((f)/(f^(\prime))-1)v \end{gathered}`

Substituting the values, the speed of the car will be

`\begin{gathered} v^(\prime)=((f)/((9)/(10)f)-1)*344.31 \\ =38.25m/s^2 \end{gathered}`