Question

Two sources of sound S1 and S2 vibrate at same frequency and are in phase. The intensity of sound detected at a point P as shown in the figure is I0. (a) If θ equals 45°, what will be the intensity of sound detected at this point if one of the sources is switched off? (b) What will be the answer of the previous part if θ = 60°?

Answer 1

a) Let the intensity at P due to one source = I and the pressure amplitude = p₀. When both sources are active the amplitude of interfering wave = 2p₀ {Since both are in phase}. The intensity is proportional to the square of the pressure amplitude, hence I/I₀ = p₀²/(2p₀)² = 1/4  →I = I₀/4   (b) Since the above result is independent of θ, so the intensity with only one source active will be the same even if θ = 60°. i.e. I = I₀/4

Answer 2

Given that,

Two sources of sound S1 and S2 vibrate at same frequency and are in phase.

The intensity of sound detected at a point P is I₀.

Suppose the amplitude is

$A_{1}=A_{2}=a$

$S_{2}P-S_{1}P=0$

(I). Foe $\theta=45^{\circ}$

The path difference is zero because both source are in phase.

So, $S_{2}P-S_{1}P=0$

We need to calculate the intensity of sound detected at this point

Using formula of intensity

$I_{max}=(A_{1}+A_{2})^2$

$I_{max}=4a^2$

The intensity at a point P is I₀

so, $I_{0}=4a^2$

$a^2=\dfrac{I_{0}}{4}$

If the one of the sources is switched off

So, the intensity of sound detected at this point is

$I_{max}=a^2$

Put the value of a into the formula

$I_{max}=\dfrac{I_{0}}{4}$

(II). If θ = 60°, then the intensity of sound detected at a point P is I₀.

We know that,

The intensity is independent of θ.

The path difference is zero because both source are in phase.

So, $S_{2}P-S_{1}P=0$

If one of the sources is switched off

We need to calculate the intensity of sound detected at this point

$I_{max}=a^2$

Put the value of a into the formula

$I_{max}=\dfrac{I_{0}}{4}$

Hence, (I). The intensity of sound detected at this point is $\dfrac{I_{0}}{4}$

(II). If θ = 60° then, the intensity of sound detected at this point is $\dfrac{I_{0}}{4}$