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Chapter 10: Problem 78

A hedge trimmer has a blade that moves back and forth with a frequency of $28 \mathrm{Hz}$. The blade motion is converted from the rotation provided by the electric motor to an oscillatory motion by means of a Scotch yoke (see Conceptual Question 7 ). The blade moves \(2.4 \mathrm{cm}\) during each stroke. Assuming that the blade moves with SHM, what are the maximum speed and maximum acceleration of the blade?

Short Answer

Expert verified
Answer: The maximum speed of the hedge trimmer's blade is approximately 4.216 m/s, and the maximum acceleration is approximately 293.65 m/s虏.

Step by step solution

01

Calculate Angular Frequency (\(\omega\))

First, we need to find the angular frequency of the oscillation. We can do this by using the formula \(\omega = 2\pi f\), where \(f\) is the given frequency. In this case, \(f = 28\,\text{Hz}\): \(\omega = 2\pi(28) = 56\pi \, \text{rad/s}\)
02

Calculate Maximum Speed (\(v_{max}\))

Next, we will find the maximum speed of the blade. The formula for maximum speed in SHM is \(v_{max} = \omega A\), where \(\omega\) is the angular frequency and \(A\) is the amplitude (in meters). The given amplitude is \(2.4\,\text{cm}\), which is equivalent to \(0.024\,\text{m}\): \(v_{max} = (56\pi) (0.024) \approx 4.216\,\text{m/s}\)
03

Calculate Maximum Acceleration (\(a_{max}\))

Finally, we will find the maximum acceleration of the blade. For SHM, the formula for maximum acceleration is \(a_{max} = \omega^2 A\). We already know \(\omega\) and \(A\): \(a_{max} = (56\pi)^2 (0.024) \approx 293.65\,\text{m/s}^2\) The maximum speed of the blade is approximately \(4.216\,\text{m/s}\), and the maximum acceleration is approximately \(293.65\,\text{m/s}^2\).

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