How much additional time will the slower pulse require to reach the end of its string?

Two strings, each 16.8 m long, are stretched side by side. One string has a mass of 78.0 g and a tension of 180.0 N. The second string has a mass of 58.0 g and a tension of 155.4 N. A pulse is generated at one end of each string simultaneously.

Once the faster pulse reaches the far end of its string, how much additional time will the slower pulse require to reach the end of its string?

Wave speed v = sqrt(T/mu) where T is the tension in N and mu is the linear mass density in kg/m

First string v = sqrt(180N/(0.078kg/16.8m)) = 196.9 m/s

Second string v = sqrt(155.4N/(0.058kg/16.8m)) = 212.2m/s

It takes the first string 16.8m/196.9m/s = 0.0853s to travel the length

It takes the second string 16.8/212.2 = 0.07917s to travel

Therefore the slower wave arrives 0.0853-0.07917 = 0.00615s later

This entry was posted on Thursday, December 3rd, 2009 at 2:47 pm and is filed under pulse. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

One Response to “How much additional time will the slower pulse require to reach the end of its string?”

Debbie on December 3rd, 2009 at 7:59 pm

Wave speed v = sqrt(T/mu) where T is the tension in N and mu is the linear mass density in kg/m

First string v = sqrt(180N/(0.078kg/16.8m)) = 196.9 m/s

Second string v = sqrt(155.4N/(0.058kg/16.8m)) = 212.2m/s

It takes the first string 16.8m/196.9m/s = 0.0853s to travel the length

It takes the second string 16.8/212.2 = 0.07917s to travel

Therefore the slower wave arrives 0.0853-0.07917 = 0.00615s later
References :

Pulse