Question

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

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

Step-by-step solution

Convert the displacement amplitude to meters

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

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.

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: \(ω^2 = \dfrac{12 \, \mathrm{m/s^2}}{0.05 \, \mathrm{m}}\) \(ω^2 = 240 \, \mathrm{s^{-2}}\) \(ω = \sqrt{240 \, \mathrm{s^{-2}}} = 15.49 \, \mathrm{rad/s}\)

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.