1. Let the three sound waves be $y_1=a\cos 2\pi (v-1)t,\quad y_2=a\cos 2\pi vt,\quad y_3=a\cos 2\pi (v+1)t.$

2. Their superposition is $y=y_1+y_2+y_3=a\left[\cos 2\pi (v-1)t+\cos 2\pi vt+\cos 2\pi (v+1)t\right].$

3. Combine the first and third terms using $\cos(A-B)+\cos(A+B)=2\cos A\cos B.$ Here, take $A=2\pi vt,\quad B=2\pi t.$ Then $\cos 2\pi (v-1)t+\cos 2\pi (v+1)t=2\cos 2\pi vt\cos 2\pi t.$

   So, $y=a\cos 2\pi vt\,[1+2\cos 2\pi t].$

4. Hence the amplitude varies as $A(t)=a[1+2\cos 2\pi t].$ The beat frequency is determined by how many times per second the loudness becomes maximum.

5. Since loudness depends on the magnitude of amplitude, consider $|A(t)|=a|1+2\cos 2\pi t|.$ Over one second, $\cos 2\pi t$ completes one full cycle.

6. Maximum loudness occurs when $|1+2\cos 2\pi t|$ is maximum. This happens at:

   • $\cos 2\pi t=1 \Rightarrow |1+2(1)|=3$
   • $\cos 2\pi t=-1 \Rightarrow |1+2(-1)|=1$

   The principal maxima of intensity occur twice in one second due to the modulation pattern of the three close frequencies.

7. Therefore, the number of beats heard per second is $2.$

8. Checking options:

   • A: $3$ ❌
   • B: $2$ ✅
   • C: $1$ ❌
   • D: $4$ ❌

Therefore, the correct answer is B.