Video: Determining the Average Power in Mechanical Waves on a String

A string of length 5.0 m and mass of 90 g is held under a tension of 1.0 ร— 10ยฒ N. A wave travels down the string that is modeled as ๐‘ฆ (๐‘ฅ, ๐‘ก) = 0.010 m sin (0.40๐‘ฅ โˆ’ 1170.12๐‘ก), where ๐‘ฅ is measured in meters and ๐‘ก is measured in seconds. What is the power of the wave over one wavelength?

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Video Transcript

A string of length 5.0 meters and mass of 90 grams is held under a tension of 1.0 times 10 to the two newtons. A wave travels down the string that is modeled as ๐‘ฆ as a function of ๐‘ฅ and ๐‘ก equals 0.010 meters times the sin of 0.40 ๐‘ฅ minus 1170.12 times ๐‘ก, where ๐‘ฅ is measured in meters and ๐‘ก is measured in seconds. What is the power of the wave over one wavelength?

Letโ€™s begin our solution by highlighting some of the vital information given. Our string has a length of 5.0 meters. Weโ€™ll refer to that as capital ๐ฟ. And it has a mass of 90 grams. Weโ€™ll refer to that as ๐‘š. The string is held under a tension of 1.0 times 10 to the two newtons. Weโ€™ll call that value capital ๐น sub ๐‘ก. Weโ€™re also given an equation, ๐‘ฆ as a function of ๐‘ฅ and ๐‘ก, which tells us the position of the string over time. We are asked to solve for the power of the wave over one wavelength. Weโ€™ll call this capital ๐‘ƒ sub avg for average power.

Letโ€™s begin our solution by looking at this equation, ๐‘ฆ as a function of ๐‘ฅ and ๐‘ก, a little more closely. Here, weโ€™ve written down everything weโ€™ve been given about the problem on the right. But we can go even further by looking into this equation, ๐‘ฆ as a function of position and time. Whenever we have an equation of this form, certain parts of the equation represent particular values of our wave.

The prefactor before the sine function, which in this case is 0.010 meters, that value is the amplitude of the wave, ๐ด. As we look farther along the equation, the value multiplying ๐‘ฅ, which in our case is 0.40, is equal to the wave number of our wave represented by ๐‘˜. In further along, the value multiplying the time ๐‘ก, which in our case is 1170.12, that is equal to the waveโ€™s angular frequency ๐œ”. So weโ€™ve now greatly increased are known information about this problem by analyzing this position and time function.

Now letโ€™s refer to a relationship between the average power of a wave on a string and the properties of that string. The average power on the string, that is the power over one wavelength, is equal to one-half times the square root of ๐œ‡, the mass of the string per unit length, multiplied by ๐‘“, the tension in the string, times ๐œ” squared multiplied by the amplitude, ๐ด, squared.

When we apply this relationship to our scenario, we can see that for each of the variables on the right side of this equation, we have a value either given in the problem statement or derived from the wave position equation, except for the variable ๐œ‡, the mass per unit length of the string. But, given that ๐œ‡ is defined as the mass of the string divided by its length, we can use the given mass and length of our string to solve for that variable. Weโ€™re now ready to begin inserting the values for our variables to the right side of our equation for average power.

Before we multiply these values together, letโ€™s convert the mass of 90 grams to a mass in kilograms. 90 grams is equal to 0.090 kilograms. Now all the values in this expression are in SI base units and ready to be multiplied.

When we enter these values on our calculator, to two significant figures, we get an average power of 92 watts. That is how much power the wave has over one wavelength.

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