91Ó°ÊÓ

The power output subjected to a particular frequency can be determined using the equation of sound level. The intensity of the sound is proportional to the logarithmic ratio of power.

Given data:

The rating of amplifier at 1000 Hz is \({p_{1000}} = 225\;{\rm{W}}\).

The level of sound is \(\beta = 12\;{\rm{dB}}\)

The value of ouput at 15 kHz can be calculated as:

\(\begin{aligned}{c}\beta &= 10\log \frac{{{p_{15}}}}{{{p_{1000}}}}\\ - 12 &= 10\log \frac{{{p_{15}}}}{{\left( {225\;{\rm{W}}} \right)}}\\ - 1.2 &= \log \frac{{{p_{15}}}}{{\left( {225\;{\rm{W}}} \right)}}\\{10^{ - 1.2}} &= \frac{{{p_{15}}}}{{\left( {225\;{\rm{W}}} \right)}}\\{p_{15}} &= 225\left( {{{10}^{ - 1.2}}} \right)\\ &= 14\;{\rm{W}}\end{aligned}\)

Thus, the ouput at 15 kHz is \(14\;{\rm{W}}\).