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Chapter 13: Problem 27

A particle undergoes simple harmonic motion with maximum speed \(1.4 \mathrm{m} / \mathrm{s}\) and maximum acceleration \(3.1 \mathrm{m} / \mathrm{s}^{2} .\) Find the (a) angular frequency, (b) period, and (c) amplitude.

Short Answer

Expert verified
The solutions are (a) angular frequency \(\omega\) can be calculated from \(\sqrt{a_{max} / v_{max}}\), (b) period can be calculated from \(2\pi / \omega\), and (c) amplitude can be calculated from \(v_{max} / \omega\) - make sure to calculate \(\omega\) first and then substitute its value into the formulas for T and A.

Step by step solution

01

Calculate the Angular Frequency (\(\omega\))

We can calculate the angular frequency (\(\omega\)) using the relationship \(a_{max} = \omega^2 A\), and solving for \(\omega\). However, we need the amplitude (A) for this. Fortunately, we can get A by using the other relationship \(v_{max} = \omega A\), to solve for A as: \(A = v_{max} / \omega\). Then substituting this into our first relationship gives us \(\omega = \sqrt{a_{max} / A} = \sqrt{a_{max} * \omega / v_{max}}\), simplifying and solving for \(\omega\) yields: \(\omega = \sqrt{a_{max} / v_{max}}\).
02

Calculate the Period (T)

After obtaining the angular frequency (\(\omega\)), we can now calculate the period of the motion (T). The relationship between angular frequency and period is given by \(\omega = 2\pi / T\). Rearranging this for T gives us: \(T = 2\pi / \omega\). Substituting \(\omega\) from the previous step allows us to find the period of the simple harmonic motion.
03

Calculate the Amplitude (A)

Finally, we can calculate the amplitude (A), which is the maximum displacement from the equilibrium position. We can use the relationship \(v_{max} = \omega A\) and solve for A like so: \(A = v_{max} / \omega\). Substituting the value of \(\omega\) from step 1 allows us to find the amplitude of the motion.

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Most popular questions from this chapter

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