Exercise 15.17: 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

0.337 kg

Explanation:

Given:

length of wire 'l'= 3.80m

mass of suspended object 'M'= 54kg

transverse pulse 't'= 0.0492 s

mass of wire 'm' =?

we can calculate mass from the linear mass density formula i.e

μ = m/l ---> eq A

where m= mass and l=length

but first we have to calculate 'μ' linear mass density

we can determine the linear mass density μ by using the formula of wave in a string speed

v = sqrt (F/μ)---> eq 1

starting with the formula of wave speed that is

v= l / t

where, 'l' is the distance traveled and 't' is time interval

v= 3.80/ 0.0492 =>77.2 m/s

next is to apply newton's second law to the hanged mass in order to find force.

ΣF = F - Fg

F= Fg => Mg

F= 54.0 x 9.8

F = 529 N

putting  the values of v and F in equation 1 we get;

77.2 = sqrt( 529/ μ) ---> taking square on both sides

μ= 529/ 5959.84

μ= 0.089kg/m

putting the values of μ and L in eq A, we get

μ= m/l

0.089= m / 3.80

m= 0.089 x 3.80

m = 0.337 kg

therefore, the mass of the wire is  0.337 kg

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