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Chapter 10: Problem 29

The prong of a tuning fork moves back and forth when it is set into vibration. The distance the prong moves between its extreme positions is $2.24 \mathrm{mm}\(. If the frequency of the tuning fork is \)440.0 \mathrm{Hz},$ what are the maximum velocity and the maximum acceleration of the prong? Assume SHM.

Short Answer

Expert verified
Answer: The maximum velocity of the tuning fork's prong is \(972.56 \cdot 10^{-3}\:\mathrm{m/s}\) and the maximum acceleration is \(28.56 \cdot 10^{3}\:\mathrm{m/s^2}\).

Step by step solution

01

Identify the given information

First, we need to note the information given to us in the problem: Amplitude (A) = half the distance between extreme positions = \(2.24\:\mathrm{mm} / 2 = 1.12\:\mathrm{mm} = 1.12 \cdot 10^{-3} \:\mathrm{m}\) Frequency (f) = \(440.0\:\mathrm{Hz}\)
02

Compute the angular frequency

To compute the angular frequency (\(\omega\)), we use the formula: \(\omega = 2\pi f\) Substitute the given frequency (f): \(\omega = 2\pi(440.0\:\mathrm{Hz}) = 880\pi\:\mathrm{rad/s}\)
03

Calculate the maximum velocity

The formula for maximum velocity (v_max) in SHM is: \(v_\max = A\omega\) Substitute the computed angular frequency and given amplitude: \(v_\max = (1.12 \cdot 10^{-3}\:\mathrm{m})(880\pi\:\mathrm{rad/s}) = 972.56 \cdot 10^{-3}\:\mathrm{m/s}\)
04

Calculate the maximum acceleration

The formula for maximum acceleration (a_max) in SHM is: \(a_\max = A\omega^2\) Substitute the computed angular frequency and given amplitude: \(a_\max = (1.12 \cdot 10^{-3}\:\mathrm{m})(880\pi\:\mathrm{rad/s})^2 = 28.56 \cdot 10^{3}\:\mathrm{m/s^2}\)
05

Write the final answer

Now we have found the maximum velocity and maximum acceleration of the tuning fork's prong: Maximum Velocity: \(v_\max = 972.56 \cdot 10^{-3}\:\mathrm{m/s}\) Maximum Acceleration: \(a_\max = 28.56 \cdot 10^{3}\:\mathrm{m/s^2}\)

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