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

Answer:

$5.87*10^{-4}W$

Explanation:

Given that:

$\beta = 89,6 dB$

$D = 13.0 cm$

$r = \frac{D}{2}\\ = \frac{13.0}{2}\\ = 6.5 cm\\=0.065m$

Efficiency = 2.06 % = 0.0206

The  intensity level of sound is given by the formula:

$\beta = (10dB) log (\frac{I}{I_o} )$

$\frac{\beta}{(10dB) } =log (\frac{I}{I_o} )$

Taking their exponential; we have :

$10^{\frac{\beta}{10dB}}= \frac{I}{I_o}$

$I = I_o(10^{\frac{\beta}{10dB}})$

Replacing our values; we have:

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

$= 9.12 *10^{-4}W/m^2$

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;

$P_{in}= \frac{P_{out}}{0.0206}$

$P_{in}= \frac{{1.2099048*10^{-5}}}{0.0206}$

$P_{in}= 5.87*10^{-4}W$

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 $5.87*10^{-4}W$