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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

An ideal spring with a spring constant of \(15 \mathrm{N} / \mathrm{m}\) is suspended vertically. A body of mass \(0.60 \mathrm{kg}\) is attached to the unstretched spring and released. (a) What is the extension of the spring when the speed is a maximum? (b) What is the maximum speed?
The leg bone (femur) breaks under a compressive force of about $5 \times 10^{4} \mathrm{N}\( for a human and \)10 \times 10^{4} \mathrm{N}$ for a horse. The human femur has a compressive strength of \(1.6 \times 10^{8} \mathrm{Pa},\) while the horse femur has a compressive strength of $1.4 \times 10^{8} \mathrm{Pa} .$ What is the effective crosssectional area of the femur in a human and in a horse? (Note: since the center of the femur contains bone marrow, which has essentially no compressive strength, the effective cross- sectional area is about \(80 \%\) of the total cross-sectional area.)
What is the maximum load that could be suspended from a copper wire of length \(1.0 \mathrm{m}\) and radius \(1.0 \mathrm{mm}\) without breaking the wire? Copper has an elastic limit of \(2.0 \times 10^{8} \mathrm{Pa}\) and a tensile strength of \(4.0 \times 10^{8} \mathrm{Pa}\).
An object of mass \(m\) is hung from the base of an ideal spring that is suspended from the ceiling. The spring has a spring constant \(k .\) The object is pulled down a distance \(D\) from equilibrium and released. Later, the same system is set oscillating by pulling the object down a distance \(2 D\) from equilibrium and then releasing it. (a) How do the period and frequency of oscillation change when the initial displacement is increased from \(D\) to $2 D ?$ (b) How does the total energy of oscillation change when the initial displacement is increased from \(D\) to \(2 D ?\) Give the answer as a numerical ratio. (c) The mass-spring system is set into oscillation a third time. This time the object is pulled down a distance of \(2 D\) and then given a push downward some more, so that it has an initial speed \(v_{i}\) downward. How do the period and frequency of oscillation compare to those you found in part (a)? (d) How does the total energy compare to when the object was released from rest at a displacement \(2 D ?\)
Show that, for SHM, the maximum displacement, velocity, and acceleration are related by \(v_{\mathrm{m}}^{2}=a_{\mathrm{m}} A\).
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