I don\'t know how to solve it with the variable (n) A rope, under a tension of 2

Question

I don't know how to solve it with the variable (n)

A rope, under a tension of 294 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by

y = (0.16 m) sin[(?x/4.0] sin[13 ?t].

where x = 0 at one end of the rope, x is in meters, and t is in seconds.

What are (a) the length of the rope,(b) the speed of the waves on the rope, and (c) the mass of the rope? (d) If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation?

Explanation / Answer

F = 294 N
y = 0.16* (sin((?*x/4))) * sin(13?*t)
y = 2A sin (k*x) * sin (? * t)

a) in a second-harmonic, the wave length = the length of the rope
k = 2?/? = ?/4 --> ? =8 m
The length of the rope L = 8 m

b) The sped of the waves
v = ?/k =13?/(?/4) = 52 m/s

c) melde's law :
v = ?(F *L/m)
--> m = F * L/v^2
=294 * 8/(52)^2
= 0.869 kg

d)in a third-harmonic :
L = 1.5 ? ---> ? =8/1.5 = 5.33 m
v = ?/T --> T = ?/v = (8/1.5)/52 = 0.102 s

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