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Chapter 13: Problem 26

A particle undergoes simple harmonic motion with amplitude \(25 \mathrm{cm}\) and maximum speed \(4.8 \mathrm{m} / \mathrm{s} .\) Find the (a) angular frequency, (b) period, and (c) maximum acceleration.

Short Answer

Expert verified
The angular frequency is \(19.2 \, s^{-1}\), the period is \(0.052 \, s\) and the maximum acceleration is \(92.16 \, m/s^2\).

Step by step solution

01

Find the angular frequency

To find the angular frequency \(\omega\) we use formula of maximum speed, \(v_{max}= \omega \cdot a\), where \(a\) is amplitude. Rearranging this equation gives us \(\omega = \frac{v_{max}}{a}\). With \(v_{max} = 4.8 \, m/s\) and \(a = 25 \, cm = 0.25 \, m\), \(\omega = \frac{4.8}{0.25} = 19.2 \, s^{-1}\).
02

Find the period

The period \(T\) is the inverse of the angular frequency, thus \(T = \frac{1}{\omega}\). With \(\omega = 19.2 \, s^{-1}\), substitute this to find \(T = \frac{1}{19.2} = 0.052 \, s\).
03

Find the maximum acceleration

The maximum acceleration \(a_{max}\) can be found using the relationship \(a_{max} = \omega^2 \cdot a\). With \(\omega = 19.2 \, s^{-1}\) and \(a = 0.25 \, m\), substitute these to find \(a_{max} = (19.2)^2 \cdot 0.25 = 92.16 \, m/s^2\).

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Most popular questions from this chapter

Show that the potential energy of a simple pendulum is proportional to the square of the angular displacement in the small-amplitude limit.

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