A small source of sound S having a frequency of 500 Hz is tied to...

At the highest point, the string just remains tight, so the tension is zero. The centripetal force is provided entirely by gravity:

$\dfrac{m v_{top}^2}{R} = m g \Rightarrow v_{top} = \sqrt{g R}$

Using conservation of mechanical energy between the highest point and the point at the same height as the centre:

$\dfrac{1}{2} m v^2 = \dfrac{1}{2} m v_{top}^2 + m g R$

$\dfrac{1}{2} m v^2 = \dfrac{1}{2} m (g R) + m g R = \dfrac{3}{2} m g R$

$v = \sqrt{3 g R}$

Substituting $g = 10 \text{ m s}^{-2}$ and $R = 1.6 \text{ m}$:

$v = \sqrt{3 \times 10 \times 1.6} = \sqrt{48} \approx 6.93 \text{ m s}^{-1}$

When the source is at the same height as the centre, its velocity is directed vertically. Since the observer is at a large distance vertically above the centre, the velocity of the source is directly towards or away from the observer.

The maximum frequency heard (when the source moves upwards, towards the observer) is:

f_\max = f_0 \left( \dfrac{v_0}{v_0 - v} \right) = 500 \left( \dfrac{330}{330 - 6.93} \right) \approx 510.7 \text{ Hz} \approx 511 \text{ Hz}

The minimum frequency heard (when the source moves downwards, away from the observer) is:

f_\min = f_0 \left( \dfrac{v_0}{v_0 + v} \right) = 500 \left( \dfrac{330}{330 + 6.93} \right) \approx 489.7 \text{ Hz} \approx 490 \text{ Hz}

Answer: $490 \text{ Hz}$ and $511 \text{ Hz}$