Answer:
[tex]f=2.236\ Hz[/tex]
Explanation:
Given:
Length of a rope,[tex]l=30\ m[/tex]
Position of Canary on the rope from one end, [tex]l_c=10\ m[/tex]
Position of Grackle on the rope from another end, [tex]l_g=5\ m[/tex]
Tension in the rope, [tex]F_T=200\ N[/tex]
linear mass distribution on the rope, [tex]\mu=0.1\ kg.m^{-1}[/tex]
We have for the speed of wave on the string:
[tex]v^2=\frac{F_T}{\mu}[/tex]
[tex]v^2=\frac{200}{\0.1}[/tex]
[tex]v=44.7\ m.s^{-1}[/tex]
For canary to be undisturbed we need a node at this location.
Also, at the end close to Canary there must be a node to avoid any change in pattern of vibration.
So,
the distance between Canary and the closer end must be equal to half the wavelength.
[tex]\frac{\lambda}{2} =10\ m[/tex]
[tex]\Rightarrow \lambda=20\ m[/tex]
∴Wavelength of wave to be produced = 20 m. This will give us nodes at the multiples of 10 and anti-nodes at the multiples of 5.
Now, frequency:
[tex]f=\frac{v}{\lambda}[/tex]
[tex]f=\frac{44.7}{20}[/tex]
[tex]f=2.236\ Hz[/tex]
