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Lynna [10]
4 years ago
8

Loudspeakers can produce intense sounds with surprisingly small energy input in spite of their low efficiencies. Calculate the p

ower 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.)
Physics
1 answer:
Dennis_Churaev [7]4 years ago
8 0

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

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