String 1 has a linear density of 2.60 g/m and string 2 has a linear density of 3
ID: 1893672 • Letter: S
Question
String 1 has a linear density of 2.60 g/m and string 2 has a linear density of 3.30 g/m. A student sends a pulse in both directions by quickly pulling on the knot then releasing it. Consider the pulses are to reach the ends of the strings simultaneously.-the knot attaches stings 1 and 2
-both strings are fixed at their opposite ends
-there entire length together is 4m
What should string 1's length be?
What should string 2's lenght be?
Explanation / Answer
Start with the fact that the time for each pulse is the same for both sections of string. Velocity for string one is: v1 = L1 / t Then using the equation for the speed of a wave on a string, we can write: L1 / t = v[F / µ1] Solved for t: t = L1 / v[F / µ1]-------------->(1) Similarly, the time for the pulse on the other part of the string is: t = L2 / v[F / µ2]-------------->(2) The length of either section of string may be expressed in terms of the other, so: L2 = 4.0m - L1----------------->(3) Setting (1) and (2) equal and subbing in (3), you get the quadratic equation (note that you have to also plug in the given values for µ1 and µ2 and that the tensions F are the same and divide out): (0.80)L1² - (28.8)L1 + 57.6 = 0 From the quadratic formula, we obtain the roots L1 = 34m and L1 = 2.1m. Obviously, 34m is too large, so L1 must be 2.1m in length. Therefore L2 is: L2 = 4.0m - L1 = 4.0m - 2.1m = 1.9m