Question

a canary sits 10 m from the edge 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 exerts 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: frequency = 2.23Hz

Step-by-step explanation:

1. The rope on which the birds stands is pulled by two poles that each exerts 200N on it. The mass per unit length = u = 0.1kg/m

2. In order to keep the canary undisturbed, a standing wave should exist on the rope while the position of the canary must be at one of the modes of the standing wave. If we started measuring the distance from the length side of the rope we can that x = 5m (Position of the grackle), the wave should have a node at x = 20m ( position of canary). We can deduce that ∆ = 2/3 × L

Therefore, ∆ = 2/3 × 30 = 20m

3. Therefore, speed of the wave on the rope using the relation V=√f/u

We substitute u = 0.1kg/m and f = 200N

V = √200N/0.1kg/m = 44.7m/s

Therefore, frequency f = V/∆ = 44.7m/s/20m = 2.23Hz