Question

Two sources of sound $S_1$ and $S_2$ vibrate at same frequency and are in phase (figure below). The intensity of sound detected at a point $P$ as shown in the figure is $I_0$. If θ equals $60°$, what will be the intensity of sound detected at this point if one of the sources is switched off?

Answer 1

Step 1: Given

Frequency of sources: $f_1=f_2=f$

Amplitude of sources: $A_1=A_2=a$

Intensity at point P: $I_0$

Angle: $\theta =60°$

Step 2: Formula Used

$I_{max}={\left(A_1+A_2\right)}^2$

Step 3: Find the intensity

Both the sources are in same phase, which implies that the path difference will be $0$. Calculate the value of a by using the formula.

$$\begin{gathered} I_{max}={\left(A_1+A_2\right)}^2 \\ \Rightarrow I_0={\left(a+a\right)}^2 \\ \Rightarrow I_0=4a^2 \\ \Rightarrow a^2=\frac{I_0}{4} \end{gathered}$$

When one of the sources is switched off, the intensity will be only due to the square of amplitude of one of the source. Calculate the intensity.

$$\begin{gathered} I=a^2 \\ =\frac{I_0}{4} \end{gathered}$$

Hence, the intensity of sound is $\frac{I_0}{4}$.