In case 1, a source of sound approaches a stationary observer at speed u. In case 2, the observer moves toward the stationary source at the same speed u. If the source is always producing the same frequency sound, will the observer hear the same frequency in both cases, since the relative speed is the same each time? Why or why not?

The frequency detected by the observer is;

$f_o=f_s\left(\frac{v_{sound}\pm v_o}{v_{sound}\mp v_s}\right)$ (1)

In the above case, because the observer is stationary $(v_o=0)$, $v_s$is the source's speed, $f_o$ is the frequency perceived by the observer, and $f_s$is the source's frequency.

Now using the minus sign in the denominator as the observer is moving toward the source.

$f_o=f_s\left(\frac{v_{sound}}{v_{sound}-v_s}\right)$ (2)

In the other case, $\left(V_0=V\right)\left(V_S=0\right)$because the source is stationary, role="math" localid="1655804494928" $\left(f_0\right)$represents the frequency detected by the observer and $\left(f_s\right)$represents the frequency of the source.

Now, using the plus sign in the denominator as the observer is moving towards the source.

$f_o=f_s\left(\frac{v_{sound}+v}{v_{sound}}\right)$ (3)

In both the above cases, the observer will not hear the same frequency as in the earlier case.

Hence, the observer will not hear the same frequency in both circumstances because $f_o$ in the first case does not equal $f_o$in the second.