Two speakers S_1 and S_2 , powered by a common amplifier, are...

The wavelength of the sound wave is given by:

$\lambda = \dfrac{v}{f} = \dfrac{330}{600} = 0.55 \text{ m}$

The distance between the two speakers $S_1$ and $S_2$ is:

$d = 1.0 - (-1.0) = 2.0 \text{ m}$

For an observer at a very large distance from the origin, the path difference $\Delta x$ at an angle $\theta$ from the X-axis is:

$\Delta x = d \sin \theta$

For a maximum (constructive interference), the path difference must be an integer multiple of the wavelength:

$\Delta x = n\lambda \implies d \sin \theta = n\lambda$

Substituting the values, we get:

$2.0 \sin \theta = n(0.55) \implies \sin \theta = 0.275 n$

Since the maximum value of $\sin \theta$ is $1$ (as \theta \to 90^{}^{{}^{\circ}} when the man moves to infinity parallel to the Y-axis):

$0.275 n \le 1 \implies n \le \dfrac{1}{0.275} \approx 3.63$

The possible non-negative integer values for $n$ are $0, 1, 2, 3$.

The man starts at a point on the X-axis, where \theta = 0^{}^{{}^{\circ}}. This corresponds to the central maximum ($n = 0$).

As he moves parallel to the Y-axis, $\theta$ increases, and he will successively encounter the maxima for $n = 1, 2, 3$.

The first maximum he encounters while walking corresponds to $n = 1$.

If he continues walking past this first maximum, the additional maxima he will hear correspond to $n = 2$ and $n = 3$.

Therefore, he will hear $2$ additional maxima.