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Chapter 10: Problem 38

A small bird's wings can undergo a maximum displacement amplitude of $5.0 \mathrm{cm}$ (distance from the tip of the wing to the horizontal). If the maximum acceleration of the wings is \(12 \mathrm{m} / \mathrm{s}^{2},\) and we assume the wings are undergoing simple harmonic motion when beating. what is the oscillation frequency of the wing tips?

Short Answer

Expert verified
Answer: The oscillation frequency of the wing tips is approximately 2.46 Hz.

Step by step solution

01

Convert the displacement amplitude to meters

The given displacement amplitude is in centimeters, so we need to convert it to meters: \(A = 5.0 \, \mathrm{cm} \times \dfrac{1 \, \mathrm{m}}{100 \, \mathrm{cm}} = 0.05 \, \mathrm{m}\)
02

Set up the formula for maximum acceleration

The formula for maximum acceleration in simple harmonic motion is: \(a = Aω^2\) We know the maximum acceleration \(a = 12 \, \mathrm{m/s^2}\) and the amplitude \(A = 0.05 \, \mathrm{m}\). Now we can plug in the values to solve for ω.
03

Solve for the angular frequency ω

Plugging the values into the equation, we get: \(12 \, \mathrm{m/s^2} = 0.05 \, \mathrm{m} \times ω^2\) Now, to find ω, we can rearrange the equation and solve for ω: \(ω^2 = \dfrac{12 \, \mathrm{m/s^2}}{0.05 \, \mathrm{m}}\) \(ω^2 = 240 \, \mathrm{s^{-2}}\) \(ω = \sqrt{240 \, \mathrm{s^{-2}}} = 15.49 \, \mathrm{rad/s}\)
04

Find the oscillation frequency using the formula \(f = \dfrac{ω}{2\pi}\)

Now, we can find the oscillation frequency using the formula: \(f = \dfrac{ω}{2\pi} = \dfrac{15.49 \, \mathrm{rad/s}}{2\pi} \approx 2.46 \, \mathrm{Hz}\) The oscillation frequency of the wing tips is approximately 2.46 Hz.

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Most popular questions from this chapter

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