# 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.

Let’s start our solution by highlighting some of the important information we’ve been given. We’re told that the length of the string is 3.00 meters, which we’ll call 𝐿. We’re also told the string has a total mass of 5.00 grams, which we’ll call 𝑚. The string is under a tension force of 500.00 newtons, which we’ll Call 𝐹 sub 𝑡. We want to know the time, we’ll call it 𝑡, for a pulse to travel the entire length of the string, and we’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 𝑣 is equal to the square root of the tension force on a string divided by its linear mass density, 𝜇. 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 𝑣 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 𝐿. So the square root of 𝐹 sub 𝑡 over 𝜇 is equal to 𝐿 divided by 𝑡. When we rearrange this equation to solve for 𝑡, we find that 𝑡, the time it takes the pulse to move down the length of the string, is equal to that length times the square root of 𝜇 over 𝐹 sub 𝑡.

Now in the problem statement, we’ve been given values for 𝐿 and 𝐹 sub 𝑡, but what about 𝜇? Recall that 𝜇 has units of kilograms per meter, and it is called the linear mass density of the string. So from 𝜇, we can substitute the mass of the string divided by its length. Algebraically, this simplifies our expression for the time 𝑡, which we can now write as the square root of 𝑚, the mass, times 𝐿, the length of the string, divided by the tension force acting on it, 𝐹 sub 𝑡.

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 𝑡 is equal, to three significant figures, to 5.48 milliseconds. That’s how long it takes a pulse to travel down the full length of the string.