A rope, under a tension of 380 N and fixed at both ends, oscillates in a second-

Question

A rope, under a tension of 380 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given below where x = 0 at one end of the rope, x is in meters, and t is in seconds..

y = ( 0.48 m)(sin((Pi x)/8)sin(14 Pi x)

(a) What is the length of the rope?

_____m

(b) What is the speed of the waves on the rope?

_____m/s

(c) What is the mass of the rope?

_____kg

(d) If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation?

_____s

Explanation / Answer

(a) The second harmonic has a node in the middle as well as at each end, so the length of the rope is then one wavelength.
A half-wavelength is when sin(*x/8) = 0, so x = 8m

So the length is L = 2*8m = 16m

(b) The radian frequency is = 14 rad/s = 2f
so f = 7 Hz
The wavespeed is c = f * wavelength = 7*16 m/s = 112 m/s

(c) c = sqrt(T/u)
mass per length is u = T / c^2 = 380/112^2 kg/m = 0.030 kg/m

So mass of rope = 0.030 kg/m * 16m = 0.48 kg

(d) New frequency = 7Hz * 3/2 = 10.5 Hz
So new period = 1/10.5 s = 0.095 s

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