Answer:
Step-by-step explanation:
Given
Natural frequency of vibration
![f=403\ Hz](https://img.qammunity.org/2021/formulas/physics/high-school/b3zqc7ynunqj8b4kxm9fh2ne117sfephua.png)
Velocity of source
![v_s=26.3\ m/s](https://img.qammunity.org/2021/formulas/physics/high-school/26gc13uydvt7mi5zsg8i8kp3gqnbpipqan.png)
Considering velocity of sound
![v=341\ m/s](https://img.qammunity.org/2021/formulas/physics/high-school/2hyxw5hcbkddttvdzxm2d9iej2lsa2097e.png)
Apparent Frequency of sound is given by Doppler effect
![f'=(v+v_o)/(v-v_s)\cdot f](https://img.qammunity.org/2021/formulas/physics/high-school/2hu6gy3bur8ukmfjl1gc24wn1j4k5cnawd.png)
where
![f=natural\ frequency](https://img.qammunity.org/2021/formulas/physics/high-school/40rrv8c3lij0621yzhr9z6etqcj30v0juv.png)
![v=velocity\ of\ sound\ wave](https://img.qammunity.org/2021/formulas/physics/high-school/uyxyzgg1xxln5tmwd8fizi16l4n090rm7r.png)
![v_o=velocity\ of\ observer](https://img.qammunity.org/2021/formulas/physics/high-school/qaj1e6dr5wcpo3ll7mpatf1md4750a0icu.png)
![v_s=velocity\ of\ source](https://img.qammunity.org/2021/formulas/physics/high-school/ml5i4lxyf86n8yoebr8czhudo4dxlk18nx.png)
![f'=apparent\ frequency](https://img.qammunity.org/2021/formulas/physics/high-school/idrq80c5mjxzqna9es83fmmerrofhxoimo.png)
here
![v_o=0](https://img.qammunity.org/2021/formulas/physics/high-school/l6xuembtjgh6wxqavbbs7ns84mxinpk5bb.png)
![f'=(341)/(341-26.3)\cdot 403=437.37 Hz](https://img.qammunity.org/2021/formulas/physics/high-school/z4396rv8il7to7qtlaprcxbsuf9po37b0y.png)
If you start singing at frequency of 403 Hz then apparent frequency to your friend
Observer is moving with 26.3 m/s and source is stationary
![f'=(v+26.3)/(v-0)\cdot 403](https://img.qammunity.org/2021/formulas/physics/high-school/6smvp5tqxtuc2oayxx2qzn54tidkxueprf.png)