1. Given data

• Source frequency: $f = 240\,\text{Hz}$
• In Case 1, source and observer move towards each other with equal speed $v$.
• Observed frequency in Case 1: $f_1 = 288\,\text{Hz}$
• In Case 2, source and observer move away from each other with the same speed $v$.
• We need to find the observed frequency $n$.

Let the speed of sound in air be $u$.

2. Use Doppler effect formula

For sound, when observer and source are moving in a medium:

$f' = f\,\frac{u \pm v_o}{u \mp v_s}$

where:

• $v_o$ = speed of observer
• $v_s$ = speed of source
• use $+$ in numerator when observer moves towards source
• use $-$ in numerator when observer moves away from source
• use $-$ in denominator when source moves towards observer
• use $+$ in denominator when source moves away from observer

Here, in both cases, $v_o = v_s = v$.

3. Case 1: Moving towards each other

So,

$288 = 240\cdot \frac{u+v}{u-v}$

Divide by $240$:

$\frac{288}{240} = \frac{u+v}{u-v}$

$\frac{6}{5} = \frac{u+v}{u-v}$

Cross-multiplying:

$6(u-v) = 5(u+v)$

$6u - 6v = 5u + 5v$

$u = 11v$

So,

$\frac{v}{u} = \frac{1}{11}$

4. Case 2: Moving away from each other

Now observer moves away from source and source also moves away from observer.

Thus,

$n = 240\cdot \frac{u-v}{u+v}$

Using $u = 11v$:

$n = 240\cdot \frac{11v-v}{11v+v}$

$n = 240\cdot \frac{10v}{12v}$

$n = 240\cdot \frac{5}{6}$

$n = 200\,\text{Hz}$

5. Final answer

$\boxed{200}$

The derived answer matches the stored correct answer.