Question

An eagle lands on a long telephone wire which begins to oscillate under its weight. It displaces 25 cm between the highest and lowest point, and completes each oscillation in 1.5 s. What is the maximum speed of the eagle? Report the value in m/s rounded to the nearest hundredth.

Answer 1

Distance between the highest and the lowest point = D = 25 cm = 0.25 m

Amplitude of the motion = A

A = D/2

A = 0.25/2

A = 0.125 m

Time period of oscillation = T = 1.5 sec

Angular frequency of the oscillation =

= 4.189 rad/s

Maximum speed of the eagle = V_max

Maximum speed in a simple harmonic motion is given by,

V_max = 0.52 m/s

Maximum speed of the eagle = 0.52 m/s