Question

A jet plane flies overhead at Mach 1.70 and at a constant altitude of 1250 m. (a) What is the angle $\mathit{\sigma }$of the shock-wave cone? (b) How much time after the plane passes directly overhead do you hear the sonic boom? Neglect the variation of the speed of sound with altitude.

Short answer

a) Angle of the shock-wave cone is $\alpha =36.{03}^c$

b) Time taken is $2.94s$

Step-by-step solution

Step 1:

Given data:

The shock wave is produced by a sound source that moves faster than sound

$\sin \alpha =\frac{v}{v_S}$

Here,

$v\to$speed of sound

$v_s\to$speed of source

localid="1664344351725" $\alpha \to$angle of shock wave

Step 2:

a) As the Mach number $=\frac{v_S}{v}=1.7$

$\begin{array}{l}sim=\frac{v}{v_S}=\frac{1}{1.7}\\ \alpha =si{n}^{-1}\left(\frac{1}{1.7}\right)=36.{03}^o\end{array}$

Hence, the angle of the shock-wave cone is$\alpha =36.{03}^{°}$

Step 3:

b) Time-taken is

$$\begin{gathered} \mathrm{Tan\alpha }=\tan 36.{03}^{°} \\ =\frac{\mathrm{d}}{{\mathrm{v}}_{{\mathrm{S}}^{}}\mathrm{t}}=0.727 \end{gathered}$$

On putting the values;

$$\begin{gathered} t=\frac{\mathrm{d}}{{\mathrm{v}}_{\mathrm{s}}\times 0.727} \\ =\frac{1250\mathrm{m}}{0.727\times 1.7\times 344\mathrm{m}/\mathrm{s}} \\ =2.94\mathrm{s} \end{gathered}$$

Hence, the time taken is $2.94\mathrm{s}$