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A massive bird lands on a taut light horizontal wire. You observe that after landing, the bird...

Question:

A massive bird lands on a taut light horizontal wire. You observe that after landing, the bird oscillates up and down with approximately simple harmonic motion with period T. You also notice that during the initial oscillations, the maximum height the bird reaches above its position at landing is equal to the distance below its landing position at which the bird finally comes to rest, when all oscillations have stopped. In terms of only T and g, the acceleration due to gravity, estimate the magnitude of the vertical velocity v0 that the bird had at the instant of landing on the wire.

Simple Harmonic Motion

When an object moves back and forth about the equilibrium position then such type of oscillation is known as the Simple Harmonic Oscillation. Here the total mechanical energy remains conserved at every point, only changes from one form to another.

Answer and Explanation:

Let us consider that the displacement on landing is "h"

Now, the amplitude of the bird would be (A) = 2h

From force balance, we can write

{eq}mg = kh \\ {/eq}

Now, taking the energy conservation

{eq}\dfrac{1}{2}(mv_{o}^{2}) + mg(3h) = \dfrac{1}{2}(k)(3h)^{2} \\ \dfrac{1}{2}(mv_{o}^{2}) + (kh)(3h) = \dfrac{1}{2}(k)(3h)^{2} \\ \dfrac{1}{2}(mv_{o}^{2}) = \dfrac{9kh^{2}}{2} - 3kh^{2} \\ \dfrac{1}{2}(mv_{o}^{2})= \dfrac{3kh^{2}}{2} \\ v_{o} = \sqrt{3}g\sqrt{\dfrac{m}{k}} \\ {/eq}

Now, we know that the time period is given by

{eq}T = 2\pi \sqrt{\dfrac{m}{k}} \\ \sqrt{\dfrac{m}{k}} = \dfrac{T}{2\pi} {/eq}

Putting the value in the above equation

{eq}v_{o} = \sqrt{3}g\sqrt{\dfrac{m}{k}} \\ v_{o} = \dfrac{\sqrt{3}Tg}{2\pi} {/eq}


Learn more about this topic:

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What is Energy Conservation? - Definition, Process & Examples

from ICSE Environmental Science: Study Guide & Syllabus

Chapter 1 / Lesson 6
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