Answer:
[tex]\mu=0.0049Kg/m[/tex]
Explanation:
When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:
[tex]f_n=\frac{nv}{2L}[/tex]
Where [tex]L[/tex] is the length of the string and [tex]v[/tex] the velocity of propagation. Use this expression to find the value of [tex]v[/tex].
[tex]f_6=\frac{6v}{2L}\\(150)=\frac{6v}{2(2)} \\150=\frac{3v}{2} \\3v=150(2)\\ v=\frac{300}{3} \\v=100m/s[/tex]
The velocity of propagation is given by the expression:
[tex]v=\sqrt{\frac{T}{\mu }[/tex]
Where [tex]\mu[/tex] is the desirable variable of the problem, the linear mass density, and [tex]T[/tex] is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:
[tex]T=W=mg=(5)(9.81)=49.05N[/tex]
With the value of the tension and the velocity you can find the mass density:
[tex]v=\sqrt{\frac{T}{\mu}[/tex]
[tex]v^2=\frac{T}{\mu}\\ \mu=\frac{T}{v^2} =\frac{49.05}{(100)^2} =\frac{49.05}{10000} =0.0049Kg/m[/tex]
