A source ($$S$$) of sound has frequency 240 Hz. When the observer ($$O$$) and the source move towards each other at a speed $$v$$ with respect to the ground (as shown in Case 1 in the figure), the observer measures the frequency of the sound to be 288 Hz. However, when the observer and the source move away from each other at the same speed $$v$$ with respect to the ground (as shown in Case 2 in the figure), the observer measures the frequency of sound to be $$n$$ Hz. The value of $$n$$ is ______.

The Doppler-shift formula for sound (air at rest) is
$$f' = f \,\frac{v \pm v_O}{\,v \mp v_S}$$
where
$$v$$ = speed of sound in air,  $$v_O$$ = speed of the observer,  $$v_S$$ = speed of the source.
Use the plus sign in the numerator when the observer moves $$\text$$it{towards} the source and the minus sign when he moves $$\text$$it{away}.
Use the minus sign in the denominator when the source moves $$\text$$it{towards} the observer and the plus sign when it moves $$\text$$it{away}.

The magnitude of the velocity of both source and observer relative to the ground is given as $$u$$ in each case.

Observer approaches the source ⇒ numerator $$v + u$$.
Source approaches the observer ⇒ denominator $$v - u$$.
Hence
$$288 = 240\;\frac{v + u}{v - u}$$

Simplify:
$$\frac{288}{240} = \frac{v + u}{v - u} \;\Longrightarrow\; 1.2(v - u) = v + u$$
$$1.2v - 1.2u = v + u$$
$$0.2v = 2.2u$$
$$u = \frac{0.2}{2.2}\,v = \frac{1}{11}\,v$$ $$-(1)$$

Observer recedes from the source ⇒ numerator $$v - u$$.
Source recedes from the observer ⇒ denominator $$v + u$$.
Therefore
$$n = 240\;\frac{v - u}{v + u}$$

Insert $$u = \frac{v}{11}$$ from $$(1)$$:
$$v - u = v - \frac{v}{11} = \frac{10v}{11}$$
$$v + u = v + \frac{v}{11} = \frac{12v}{11}$$

Thus
$$n = 240\;\frac{\tfrac{10v}{11}}{\tfrac{12v}{11}} = 240\;\frac{10}{12} = 240 \times \frac{5}{6} = 200$$

Therefore, the observed frequency in Case 2 is 200 Hz.