Question

Laura and Elana are discussing how to solve the following problem: "A canary sits 10 m from the end of a 30-m-long clothesline, and a grackle sits 5 m from the other end. The rope is pulled by two poles that each exert a 200-N force on it. The mass per unit length is 0.10 kg/m. At what frequency must you vibrate the line in order to dislodge the grackle while allowing the canary to sit undisturbed?"

Answer 1

Answer:

$f=2.236\ Hz$

Explanation:

Given:

Length of a rope,$l=30\ m$

Position of Canary on the rope from one end, $l_c=10\ m$

Position of Grackle on the rope from another end, $l_g=5\ m$

Tension in the rope, $F_T=200\ N$

linear mass distribution on the rope, $\mu=0.1\ kg.m^{-1}$

We have for the speed of wave on the string:

$v^2=\frac{F_T}{\mu}$

$v^2=\frac{200}{\0.1}$

$v=44.7\ m.s^{-1}$

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.

$\frac{\lambda}{2} =10\ m$

$\Rightarrow \lambda=20\ m$

∴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:

$f=\frac{v}{\lambda}$

$f=\frac{44.7}{20}$

$f=2.236\ Hz$