Comments on: How much additional time will the slower pulse require to reach the end of its string? http://www.thepulsartriyo.com/pulse/how-much-additional-time-will-the-slower-pulse-require-to-reach-the-end-of-its-string/ Mon, 21 May 2012 13:10:59 +0000 hourly 1 http://wordpress.org/?v=3.2.1 By: Debbie http://www.thepulsartriyo.com/pulse/how-much-additional-time-will-the-slower-pulse-require-to-reach-the-end-of-its-string/#comment-2658 Debbie Fri, 04 Dec 2009 00:59:59 +0000 http://www.thepulsartriyo.com/pulse/how-much-additional-time-will-the-slower-pulse-require-to-reach-the-end-of-its-string#comment-2658 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<br><b>References : </b><br> 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 :

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