A heavy rope 6.00 m long and weighing 29.4 N is attached at one end to a ceiling and hangs vertically. A 0.500-kg mass is suspended from the lower end of the rope. What is the speed of transverse waves on the rope at the (a) bottom of the rope, (b) middle of the rope, and (c) top of the rope? (d) Is the tension in the middle of the rope the average of the tensions at the top and bottom of the rope? Is the wave speed at the middle of the rope the average of the wave speeds at the top and bottom? Explain.

Short Answer

Expert verified

The speed of the transverse waves at the (a) bottom = 3.13 m/s, (b) middle = 6.26 m/s, (c) top = 8.28 m/s. The tension in the middle of the rope is the average of the tensions at the top and bottom of the rope. The wave speed at the middle of the rope is not the average of the wave speeds at the top and bottom.

Step by step solution

01

Determination of the theoretical formula of Mechanical Waves

Figures Ia, lb, and lc show the free-body diagrams. Depending on where the tension of the rope is considered, the tension in the rope assumes different values. Corresponding to each tension T there is a wave speed v given by the dynamic relationship

v=Fμ,the target variables in (a), (b) and (c).

In (d), compare the tension in the middle of the rope to the average of the tensions at the top and bottom of the rope as well as the wave speed at the middle of the rope to the average of the wave speeds at the top and bottom.

Due to the rope's weight, its tension is greater at the top Ftop than at the middle Fmiddle than at the bottom Fbottom. Hence, the wave speed increases as a wave travels up the rope. Vtop>vmiddle>vbottom, as in parts (a), (b), and (c). As of part (d), because the rope has a uniform mass, the tension in the rope increases linearly as the wave moves up along the rope, which explains why Fmiddle = Favg. However, this is not true for the wave speed since the relationship between v and F is not linear.

02

Application of the theoretical formula of Mechanical Waves

Denote the total mass of the rope bymrope.Expressing the rope's mass asmrope=wg,where w is the total weight of the rope, the rope's linear mass density is given by

μ=mropeL=wLg

The tension role="math" localid="1664341025050" Fbottom at the bottom of the rope is due to the hanging mass m. (Fig. la):

Fbottom=mg

Hence, the corresponding wave speed is

vbottom=Fbottomμ=mgw/Lg=gmLw

Substitute the known values of L, w, and m:
vbottom=Fbottomμ=9.80m/s20.500kg6.00m29.4N=3.13m/s

The tensionFmiddle at the middle of the rope is due to the hanging mass m in addition to the lower half of the rope w/2 (Fig. 1b):

Fmiddle=mg+w/2
Hence, the corresponding wave speed is
vmiddle=Fmiddleμ=mg+w/2w/Lg=Lgmgw+12


Substitute the known values of L, w, and m:

vmiddle=Fmiddleμ=6.00m9.80m/s20.500kg6.00m29.4N+12=6.26m/s

The tensionFtop at the top of the rope is due to the hanging mass m plus the total weight w of the rope (Fig. Ic):

Ftop=mg+w

Hence, the corresponding wave speed

vtop=Ftopμ=mg+ww/Lg=Lgmgw+1

Substituting the known values of L, w, and m, we find

vtop=Ftopμ=6.00m9.80m/s20.500kg6.00m29.4N+1=8.28m/s

03

Determination of the average tension and wave speed

The average of the tensions at the top and bottom of the rope is

Favg=Ftop+Fbottom2=mg+w+mg2=mg+w2

which is exactly the tensionFbottom in the middle of the rope. However, the average of the wave speeds at the top

vavg=vtop+vbottom2=8.28+3.13m/s2=5.17m/s


which is not equal to the wave speed at the middle of the rope vmiddle=6.26m/s.

Therefore, the speed of the transverse waves at the (a) bottom = 3.13 m/s, (b) middle = 6.26 m/s, (c) top = 8.28 m/s. The tension in the middle of the rope is the average of the tensions at the top and bottom of the rope.The wave speed at the middle of the rope is not the average of the wave speeds at the top and bottom.

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