Answer:
(a) [tex]f_{5}=700Hz[/tex]
(b) [tex]T=56.45N[/tex]
Explanation:
For part (a)
The wave with three antinodes has m=3.Thus f₃=3f₁ and so the fundamental frequency of string is:
[tex]f_{3}=3f_{1}\\f_{1}=\frac{f_{3}}{3}\\ f_{1}=\frac{420Hz}{3}\\ f_{1}=140Hz[/tex]
Thus the frequency of fifth harmonic is
[tex]f_{5}=5f_{1}\\f_{5}=5*140Hz\\f_{5}=700Hz[/tex]
For Part (b)
The speed of transverse wave on this string is related to the fundamental frequency by:
[tex]f_{1}=\frac{v}{2l}\\ v=2lf_{1}\\v=2(0.6m)(140Hz)\\v=168m/s[/tex]
As the speed of wave is related to tension of string and its linear density by:
[tex]v=\sqrt{\frac{T}{u} } \\T=v^{2}u\\T=(168m/s)^{2}(0.0020kg/m)\\T=56.45N[/tex]
