Q57E - University Physics with Modern Physics

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.

Step-by-Step Solution

Answer

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

b) Time taken is $2.94s$

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

$\alpha \to$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 $2.94\mathrm{s}$