Question

Two birds sit on a telephone wire, as shown in figure below. The wire is rigidly attached to the poles at A and B.a
person at pole A wishes to dislodge bird 2 without disturbing bird 1. He sets up a standing wave with a velocity o
48 m/s. At what frequency should he shake the wire to do so
(x=12m) (x=30m)
​

Answer 1

Answer:

Given:

l  = 1.5 m   , mass = 12 g

Mass of unit length = 1.512=8×10−3kg/m

T=9×g=90N

λ=1.5m, 

f=l1μT=1⋅511290000×1⋅5

=70Hz

Answer 2

Explanation:

We use Newton’s second law with the forces in the x and y directions in equilibrium.

(a) At the point where the bird is perched, the wire’s midpoint, the forces acting on the wire are the tension forces and the force of gravity acting on the bird. These forces are shown below.

(b) The mass of the bird is m=1.00kg, so the force of gravity on the bird, its weight, is mg=(1.00kg)(9.80m/$s^{2}$

)=9.80N. To calculate the angle α in the free-body diagram, we note that the base of the triangle is 25.0m, so that

   tanα= 25.0m 0.200m

​→α=0.458 0

Each of the tension forces has x and y components given by

 Tx =T*cosα and

T y =T*sinα

The x components of the two tension forces cancel out. In the y direction,

∑F y =2Tsinα−mg=0

which gives

T =  2sinαmg

  = 2sin0.458 09.80N

  =613N