Solution:

Given,
Speed of the sound (v) = 330 m/s
Frequency of source (f) =- 500 Hz

Now, for the source moving towards the stationary observer with a speed of 30 m/s. we have,

u_s = 30 m/s
u_o = 0 m/s

Then using,

\(f'=\frac v{v-u_s}\cdot f=\frac{330}{330-30}\times500=550\)

Hence, the required frequency, in this case, is 550 Hz.

Again, the observer is moving towards the stationary source with a speed of 30 m/s, we have,

us = 30 m/s
u_o = 0 m/s

Then using,

\(f'=\frac{v+u_o}v\cdot f=\frac{330+30}0\times500=545.45\) Hz.

Hence the required frequency, in this case, is 545.45 Hz

Again, for Both the source and the observer move with the speed of 30 m/s and approach one another, we have

u_s = 0 m/s
u_o = 30 m/s

The using, \(f'=\left(\frac{v+u_0}{v-u_s}\right)\times f=\left(\frac{330+30}{330-30}\right)\times500=600\) Hz

Hence, the required frequency, in this case, is 600 Hz.