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Chapter 10: Problem 28

A sewing machine needle moves with a rapid vibratory motion, rather like SHM, as it sews a seam. Suppose the needle moves \(8.4 \mathrm{mm}\) from its highest to its lowest position and it makes 24 stitches in 9.0 s. What is the maximum needle speed?

Short Answer

Expert verified
Answer: The maximum speed of the sewing machine needle is 0.1408 m/s.

Step by step solution

01

Understanding the given information

We are given the following information: - Amplitude (A): The distance from the highest to the lowest position, which is \(8.4 \mathrm{mm}\) - Number of stitches: 24 - Time taken to perform the stitches (t): 9.0 s
02

Calculate the time period for one full oscillation

One full oscillation occurs when the needle completes one up-down cycle. The time period for one full oscillation (T) can be calculated by dividing the total time taken by the number of stitches as each stitch corresponds to a single oscillation: \(T=\frac{t}{\text{number of stitches}}\) Plug in the values: \(T = \frac{9.0 \mathrm{s}}{24}\) \(T = 0.375 \mathrm{s}\)
03

Determine the angular frequency

The angular frequency (\(\omega\)) can be calculated using the formula: \( \omega = \frac{2 \pi}{T} \) Plug in the value for T: \( \omega = \frac{2 \pi}{0.375 \mathrm{s}} \) \( \omega = 16.76 \ \text{s}^{-1} \)
04

Calculate the maximum needle speed

The maximum speed (vmax) of the needle in SHM can be found using the formula: \( \mathrm{v_{max}} = A \omega \) Plug in the values for A and \(\omega\): \( \mathrm{v_{max}} = (8.4 \mathrm{mm})(16.76 \ \text{s}^{-1})\) \( \mathrm{v_{max}} = 140.8 \ \text{mm} \ \text{s}^{-1} \)
05

Express the answer in the right unit

The maximum needle speed can be converted from \(\text{mm} \ \text{s}^{-1}\) to \(\text{m} \ \text{s}^{-1}\): \( \mathrm{v_{max}} = 140.8 \ \text{mm} \ \text{s}^{-1} \times \frac{1 \text{m}}{1000 \text{mm}} = 0.1408 \ \text{m} \ \text{s}^{-1}\) The maximum needle speed is \(0.1408 \ \text{m} \ \text{s}^{-1}\).

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Most popular questions from this chapter

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