# Video: Determining the Density of a String Carrying a Wave Generated Electromechanically

A wave on a string is driven by a string vibrator which oscillates at a frequency of 100.00 Hz and an amplitude of 1.00 cm. The string vibrator operates at a voltage of 12.00 V and a current of 0.200 A. The power consumed by the string vibrator is π = πΌπ. Assume that the string vibrator is 90.0% efficient at converting electrical energy into the energy associated with the vibrations of the string. The string is 3.00 m long and is under a tension of 60.00 N. What is the linear mass density of the string?

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### Video Transcript

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. The string vibrator operates at a voltage of 12.00 volts and a current of 0.200 amps. The power consumed by the string vibrator is π equals πΌ times π. Assume that the string vibrator is 90.0-percent efficient at converting electrical energy into the energy associated with the vibrations of the string. The string is 3.00 meters long and is under a tension of 60.00 newtons. What is the linear mass density of the string?

In this statement, weβre told that the frequency of the waves on the string is 100.00 hertz; weβll call that π. Weβre told also that the amplitude of the waves on the string is 1.00 centimeters; weβll call that π΄. The voltage supplied to the string vibrator is 12.00 volts, which weβll call π. And the current supplied to the vibrator is 0.200 amps, which weβll label πΌ. The stringβs length is 3.00 meters, which weβll call πΏ.

And weβre told that the string is under a tension of 60.00 newtons; weβll call that πΉ sub π‘. We want to solve for the linear mass density of the string, which weβll call π. To begin our solution, letβs draw a diagram of the scenario. 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 π and our current πΌ.

Weβre told that the vibrator is 90-percent efficient at converting electrical power through the wire to mechanical power of the stringβs motion. We can call that efficiency, πΈ. So 90 percent of the power from the outlet is transferred to power in the string. At this point, itβs helpful to recall a relationship for power in waves on a string. The power, π, is equal to one-half the square root of π, the linear mass density, times the tension force in the string multiplied by its angular frequency, π, squared times its amplitude squared.

When we write this equation for our scenario, we see that we like to rearrange to solve for π, the linear mass density. If we multiply both sides of this equation by two divided by the square root of πΉ sub π‘ times π squared π΄ squared, then we find on the right-hand side of our equation that all terms cancel except for the square root of π. With that term isolated, if we then square both sides, then on the right-hand side, we achieve the linear mass density π by itself.

We can simplify this equation to write π is equal to four times π squared, where π is the power supply to the string divided by the tension force multiplied by π, the angular frequency, times the amplitude π΄ to the fourth power. Letβs start plugging in for these values. Weβre told in the problem statement that power, π, in an electric circuit equals the current πΌ times the voltage π.

So in our scenario, π equals πΌ times π or 0.200 amps times 12.00 volts. But thereβs one more term we want to consider. Remember that we donβt perfectly convert the electrical power to mechanical power; there is an efficiency loss, and were told that our efficiency is 90 percent. So to find the power, π, in the string which we want, weβll include our efficiency in this multiple: 90 percent or 0.90.

We can now insert this value for π in our equation for π. Next, looking at the denominator, we want to find the terms πΉ sub π‘, π, and π΄. πΉ sub π‘ and π΄ are given in the problem statement. π, the angular frequency, is equal to two π times the frequency π, which is given to us in the problem statement.

So we can plug in for those three values now: πΉ sub π‘ is 60.00 Newtons; π is 100.00 hertz; and π΄, in meters, is 0.01. When we enter these values on our calculator, we find that π is equal to 2.00 times 10 to the negative fourth kilograms per meter. Thatβs the linear mass density of the string.