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 𝑃 = 𝐼𝑉. 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 𝑃 equals 𝐼 times 𝑉. 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, we’re told that the frequency of the waves on the string is 100.00 hertz; we’ll call that 𝑓. We’re told also that the amplitude of the waves on the string is 1.00 centimeters; we’ll call that 𝐴. The voltage supplied to the string vibrator is 12.00 volts, which we’ll call 𝑉. And the current supplied to the vibrator is 0.200 amps, which we’ll label 𝐼. The string’s length is 3.00 meters, which we’ll call 𝐿.

And we’re told that the string is under a tension of 60.00 newtons; we’ll call that 𝐹 sub 𝑑. We want to solve for the linear mass density of the string, which we’ll call πœ‡. To begin our solution, let’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 𝑉 and our current 𝐼.

We’re told that the vibrator is 90-percent efficient at converting electrical power through the wire to mechanical power of the string’s motion. We can call that efficiency, 𝐸. So 90 percent of the power from the outlet is transferred to power in the string. At this point, it’s helpful to recall a relationship for power in waves on a string. The power, 𝑃, is equal to one-half the square root of πœ‡, the linear mass density, times the tension force in the string multiplied by its angular frequency, πœ”, squared times its amplitude squared.

When we write this equation for our scenario, we see that we like to rearrange to solve for πœ‡, the linear mass density. If we multiply both sides of this equation by two divided by the square root of 𝐹 sub 𝑑 times πœ” squared 𝐴 squared, then we find on the right-hand side of our equation that all terms cancel except for the square root of πœ‡. With that term isolated, if we then square both sides, then on the right-hand side, we achieve the linear mass density πœ‡ by itself.

We can simplify this equation to write πœ‡ is equal to four times 𝑃 squared, where 𝑃 is the power supply to the string divided by the tension force multiplied by πœ”, the angular frequency, times the amplitude 𝐴 to the fourth power. Let’s start plugging in for these values. We’re told in the problem statement that power, 𝑃, in an electric circuit equals the current 𝐼 times the voltage 𝑉.

So in our scenario, 𝑃 equals 𝐼 times 𝑉 or 0.200 amps times 12.00 volts. But there’s one more term we want to consider. Remember that we don’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, 𝑃, in the string which we want, we’ll include our efficiency in this multiple: 90 percent or 0.90.

We can now insert this value for 𝑃 in our equation for πœ‡. Next, looking at the denominator, we want to find the terms 𝐹 sub 𝑑, πœ”, and 𝐴. 𝐹 sub 𝑑 and 𝐴 are given in the problem statement. πœ”, the angular frequency, is equal to two πœ‹ times the frequency 𝑓, which is given to us in the problem statement.

So we can plug in for those three values now: 𝐹 sub 𝑑 is 60.00 Newtons; 𝑓 is 100.00 hertz; and 𝐴, in meters, is 0.01. When we enter these values on our calculator, we find that πœ‡ is equal to 2.00 times 10 to the negative fourth kilograms per meter. That’s the linear mass density of the string.

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