91影视

1. The power of the spherical loudspeaker is $P_s=10\text{鈥塛}$
2. The point of intensity from source of sound is$d_1=4.0\text{鈥尘}$
3. The point of intensity from source of sound is $d_2=3.0\text{鈥尘}$

Use the concept of variation of intensity with distance. The intensity of the sound from an isotropic point source decreases with the square of the distance from the source. The intensity is proportional to the square of the amplitude.

Formulae:

The intensity of the wave, $I=\frac{P_s}{4蟺r^2}$ (i)

The intensity of a wave is directly proportional to the square of the amplitude,

$I鈭�A^2$ (ii)

Using equation (i), the intensity of the sound wave at 3.0 m is given as:

$\begin{array}{c}I=\frac{10W}{4脳3.14脳{(3.0\text{m})}^2}\\ =0.088\frac{\text{W}}{{\text{m}}^{\text{2}}}\end{array}$

Hence, the value of the intensity of the sound is$0.088\frac{\text{W}}{{\text{m}}^{\text{2}}}$

The intensity of sound is proportional to the square of the amplitude. The intensity of sound at distance $d_1=4.0m$ is $I_1$ and wave amplitude is $A_1$. Hence, using equation (ii), the intensity relation is given as:

$I_1鈭�A_1^2$, $I_2鈭�A_2^2$ 鈥�.. (1)

The intensity of sound at distance $d_2=3.0m$is $I_2$ and wave amplitude is $A_2$. Hence, using equation (ii), the intensity relation is given as

$I_2鈭�A_2^2$ 鈥�.. (2)

Divide equation (1) by equation (2), we get

$\begin{array}{c}\frac{I_1}{I_2}=\frac{A_1^2}{A_2^2}\\ \frac{A_1}{A_2}=\sqrt{\frac{I_1}{I_2}}\end{array}$ 鈥�.. (3)

Using equation (i), the equation (3) becomes:

$\begin{array}{c}\frac{A_1}{A_2}=\sqrt{\frac{\left(\frac{P_s}{4蟺r_1^2}\right)}{\left(\frac{P_s}{4蟺r_2^2}\right)}}\\ =\frac{r_2}{r_1}\\ =\frac{3.0m}{4.0m}\\ =0.75\end{array}$

Hence, the value of the ratio of amplitudes is $0.75$.