Question

A 61.2 g wire is held under a tension of 170 N with one end at x = 0 and the other at x = 12.8 m. At time t = 0, pulse 1 is sent along the wire from the end at x = 12.8 m. At time t = 36.4 ms, pulse 2 is sent along the wire from the end at x = 0. At what position x do the pulses begin to meet?

Answer 1

T = tension force = 170 N

m = mass density = Total mass /Total length = 0.0612 /12.8 = 0.00478 kg/m

speed is given as

V = sqrt(T/m) = sqrt(170/0.00478) = sqrt(35564.85) = 188.6 m/s

D = distance travelled by pulse 1 in 36.4 msec = 188.6 x 0.0364 = 6.87 m

t = time for pulse 1 to reach x = 12 m   , = 12/188.6 = 63.6 ms

X' = position of pulse 2 when pulse one is at x = 12 , = 188.6 (0.0636 - 0.0364) = 5.13 m

distance between the pulses = 12 - 5.13 = 6.87

t = time to meet = 6.87/(188.6 + 188.6) = 0.0182 sec

position x where pulses meet is given as

x = 5.13 + (188.6 x 0.0182) = 8.56 m