Question

The upper end of a 3.80-m-long steel wire is fastened to the ceiling, and a 54.0-kg object is suspended from the lower end of the wire. you observe that it takes a transverse pulse 0.0492 s to travel from the bottom to the top of the wire. what is the mass of the wire?

Answer 1

Given that the lenth write steel l =3.80m fastened celling mass m= 54.0kg  
v = s/t 
 v = 3.8/0.0492 
 v = 77.23 m/s  
 now, the formula for speed in a wire is 
 v = ( T/ÎĽ )^1/2  
 where T is the tension in the string and ÎĽ is the mass pee unit length.  
 v = ( Tl/m )^1/2 .......(replace ÎĽ with m/l) 
 now T = 550N  
 v = [ 550(3.8)/m ]^1/2 
 squaring both sides  
 v^2 = 550(3.8)/m 
 v^2 = 2090/m 
 putting the value of v calculated above and solving 
  m =0.35 kg