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A wave on a string is driven by a string vibrator which oscillates at a frequency of 100.00 hertz and an amplitude of 1.00 centimeters.
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The string vibrator operates at a voltage of 12.00 volts and a current of 0.200 amps.
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The power consumed by the string vibrator is ๐ equals ๐ผ times ๐.
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Assume that the string vibrator is 90.0-percent efficient at converting electrical energy into the energy associated with the vibrations of the string.
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The string is 3.00 meters long and is under a tension of 60.00 newtons.
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What is the linear mass density of the string?
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In this statement, weโre told that the frequency of the waves on the string is 100.00 hertz; weโll call that ๐.
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Weโre told also that the amplitude of the waves on the string is 1.00 centimeters; weโll call that ๐ด.
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The voltage supplied to the string vibrator is 12.00 volts, which weโll call ๐.
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And the current supplied to the vibrator is 0.200 amps, which weโll label ๐ผ.
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The stringโs length is 3.00 meters, which weโll call ๐ฟ.
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And weโre told that the string is under a tension of 60.00 newtons; weโll call that ๐น sub ๐ก.
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We want to solve for the linear mass density of the string, which weโll call ๐.
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To begin our solution, letโs draw a diagram of the scenario.
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In our situation we have a string stretched between two ends, and one of the ends is vibrated by a string vibrator powered by our voltage ๐ and our current ๐ผ.
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Weโre told that the vibrator is 90-percent efficient at converting electrical power through the wire to mechanical power of the stringโs motion.
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We can call that efficiency, ๐ธ.
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So 90 percent of the power from the outlet is transferred to power in the string.
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At this point, itโs helpful to recall a relationship for power in waves on a string.
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The power, ๐, is equal to one-half the square root of ๐, the linear mass density, times the tension force in the string multiplied by its angular frequency, ๐, squared times its amplitude squared.
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When we write this equation for our scenario, we see that we like to rearrange to solve for ๐, the linear mass density.
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If we multiply both sides of this equation by two divided by the square root of ๐น sub ๐ก times ๐ squared ๐ด squared, then we find on the right-hand side of our equation that all terms cancel except for the square root of ๐.
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With that term isolated, if we then square both sides, then on the right-hand side, we achieve the linear mass density ๐ by itself.
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We can simplify this equation to write ๐ is equal to four times ๐ squared, where ๐ is the power supply to the string divided by the tension force multiplied by ๐, the angular frequency, times the amplitude ๐ด to the fourth power.
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Letโs start plugging in for these values.
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Weโre told in the problem statement that power, ๐, in an electric circuit equals the current ๐ผ times the voltage ๐.
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So in our scenario, ๐ equals ๐ผ times ๐ or 0.200 amps times 12.00 volts.
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But thereโs one more term we want to consider.
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Remember that we donโt perfectly convert the electrical power to mechanical power; there is an efficiency loss, and were told that our efficiency is 90 percent.
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So to find the power, ๐, in the string which we want, weโll include our efficiency in this multiple: 90 percent or 0.90.
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We can now insert this value for ๐ in our equation for ๐.
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Next, looking at the denominator, we want to find the terms ๐น sub ๐ก, ๐, and ๐ด.
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๐น sub ๐ก and ๐ด are given in the problem statement.
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๐, the angular frequency, is equal to two ๐ times the frequency ๐, which is given to us in the problem statement.
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So we can plug in for those three values now: ๐น sub ๐ก is 60.00 Newtons; ๐ is 100.00 hertz; and ๐ด, in meters, is 0.01.
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When we enter these values on our calculator, we find that ๐ is equal to 2.00 times 10 to the negative fourth kilograms per meter.
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Thatโs the linear mass density of the string.