Question

Loudspeakers can produce intense sounds with surprisingly small energy input in spite of their low efficiencies. Calculate the power input needed (in W) to produce a 89.6 dB sound intensity level for a 13.0 cm diameter speaker that has an efficiency of 2.06%. (This value is the sound intensity level right at the speaker.)

Answer 1

`5.87*10^(-4)W`

Step-by-step explanation:

Given that:

`\beta = 89,6 dB`

`D = 13.0 cm`

`r = (D)/(2)\\ = (13.0)/(2)\\ = 6.5 cm\\=0.065m`

Efficiency = 2.06 % = 0.0206

The intensity level of sound is given by the formula:

`\beta = (10dB) log ((I)/(I_o) )`

`(\beta)/((10dB) ) =log ((I)/(I_o) )`

Taking their exponential; we have :

`10^{(\beta)/(10dB)}= (I)/(I_o)`

`I = I_o(10^{(\beta)/(10dB)})`

Replacing our values; we have:

`I = (10^(-12)W/m^2)(10^{(89.6)/(10dB)})`

`= 9.12 *10^(-4)W/m^2`

Power Output;

`P_(out) = IA\\P_(out)=I(\pi r^2)\\P_(out)=(9.12*10^(-4)W/m^2)(3.14)(0.065m)^2\\P_(out)=1.2099048*10^(-5) W`

The power output that is required to produce a 89.6 dB sound intensity level for a 13.0 cm diameter speaker that has an efficiency of 2.06% is;

`P_(in)= (P_(out))/(0.0206)`

`P_(in)= \frac{{1.2099048*10^(-5)}}{0.0206}`

`P_(in)= 5.87*10^(-4)W`

Therefore, The power output that is required to produce a 89.6 dB sound intensity level for a 13.0 cm diameter speaker that has an efficiency of 2.06% is
`5.87*10^(-4)W`