Question Video: Determining the Travel Time for a Wave Pulse along a String

A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string? Give your answer in milliseconds.

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Video Transcript

A string is 3.00 meters long with a mass of 5.00 grams. The string is held taut with a tension of 500.00 newtons applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 meters of the string? Give your answer in milliseconds.

Letโs start our solution by highlighting some of the important information weโve been given. Weโre told that the length of the string is 3.00 meters, which weโll call ๐ฟ. Weโre also told the string has a total mass of 5.00 grams, which weโll call ๐. The string is under a tension force of 500.00 newtons, which weโll Call ๐น sub ๐ก. We want to know the time, weโll call it ๐ก, for a pulse to travel the entire length of the string, and weโll give our answer in milliseconds.

We can move toward our solution by recalling an equation for wave speed that depends on string tension and string mass per unit length: wave speed ๐ฃ is equal to the square root of the tension force on a string divided by its linear mass density, ๐. When we apply that relationship to our situation, we can connect it with another relationship for speed. Recall that, in general, the speed of an object ๐ฃ is equal to the distance it travels divided by the time it takes to travel that distance.

In our case, the distance the pulse travels is the length of the string ๐ฟ. So the square root of ๐น sub ๐ก over ๐ is equal to ๐ฟ divided by ๐ก. When we rearrange this equation to solve for ๐ก, we find that ๐ก, the time it takes the pulse to move down the length of the string, is equal to that length times the square root of ๐ over ๐น sub ๐ก.

Now in the problem statement, weโve been given values for ๐ฟ and ๐น sub ๐ก, but what about ๐? Recall that ๐ has units of kilograms per meter, and it is called the linear mass density of the string. So from ๐, we can substitute the mass of the string divided by its length. Algebraically, this simplifies our expression for the time ๐ก, which we can now write as the square root of ๐, the mass, times ๐ฟ, the length of the string, divided by the tension force acting on it, ๐น sub ๐ก.

When we plug in the values given for these three variables, converting the mass in grams to a mass in kilograms and plugging these values into our calculator, we find that the time ๐ก is equal, to three significant figures, to 5.48 milliseconds. Thatโs how long it takes a pulse to travel down the full length of the string.