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Chapter 14: Problem 53

A 3.0 -kg mass is vibrating on a spring. It has a resonant angular speed of \(2.4 \mathrm{rad} / \mathrm{s}\) and a damping angular speed of \(0.14 \mathrm{rad} / \mathrm{s}\). If the driving force is \(2.0 \mathrm{~N}\), find the maximum amplitude if the driving angular speed is (a) \(1.2 \mathrm{rad} / \mathrm{s},\) (b) \(2.4 \mathrm{rad} / \mathrm{s}\), and \((\) c) \(4.8 \mathrm{rad} / \mathrm{s}\).

Short Answer

Expert verified
Answer: The maximum amplitudes for the given driving angular speeds are approximately 0.187 m for 1.2 rad/s, 0.924 m for 2.4 rad/s, and 0.083 m for 4.8 rad/s.

Step by step solution

01

Identify the equation to use for maximum amplitude

The equation for the maximum amplitude of a damped harmonically driven mass on a spring is as follows: \(A_{max} = \frac{F_0}{m \sqrt{(\omega_0^2 - \omega^2)^2 + (2\beta\omega)^2}}\) Here, \(F_0\) is the driving force, \(m\) is the mass, \(\omega_0\) is the resonant angular speed, \(\omega\) is the driving angular speed, and \(\beta\) is the damping angular speed.
02

Plug in the given values

We have been given: - Mass, m = 3.0 kg - Resonant angular speed, \(\omega_0 = 2.4 \, \mathrm{rad/s}\) - Damping angular speed, \(\beta = 0.14 \, \mathrm{rad/s}\) - Driving force, \(F_0 = 2.0 \, \mathrm{N}\) We need to find the maximum amplitude for three driving angular speeds: 1. \(\omega = 1.2 \, \mathrm{rad/s}\) 2. \(\omega = 2.4 \, \mathrm{rad/s}\) 3. \(\omega = 4.8 \, \mathrm{rad/s}\)
03

Calculate the maximum amplitude for each driving angular speed

Now we can plug in the values and calculate the maximum amplitude for each driving angular speed: 1. For \(\omega = 1.2 \, \mathrm{rad/s}\): \(A_{max} = \frac{2.0}{3.0 \sqrt{({2.4}^2 - {1.2}^2)^2 + (2 \times 0.14 \times 1.2)^2}}\) \(A_{max} \approx 0.187 \, \mathrm{m}\) 2. For \(\omega = 2.4 \, \mathrm{rad/s}\): \(A_{max} = \frac{2.0}{3.0 \sqrt{({2.4}^2 - {2.4}^2)^2 + (2 \times 0.14 \times 2.4)^2}}\) \(A_{max} \approx 0.924 \, \mathrm{m}\) 3. For \(\omega = 4.8 \, \mathrm{rad/s}\): \(A_{max} = \frac{2.0}{3.0 \sqrt{({2.4}^2 - {4.8}^2)^2 + (2 \times 0.14 \times 4.8)^2}}\) \(A_{max} \approx 0.083 \, \mathrm{m}\)
04

Conclude the solution

Hence, the maximum amplitude for the given driving angular speeds are: 1. \(1.2 \, \mathrm{rad/s}\): \(\approx 0.187 \, \mathrm{m}\) 2. \(2.4 \, \mathrm{rad/s}\): \(\approx 0.924 \, \mathrm{m}\) 3. \(4.8 \, \mathrm{rad/s}\): \(\approx 0.083 \, \mathrm{m}\)

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Most popular questions from this chapter

A Foucault pendulum displayed in a museum is typically quite long, making the effect easier to see. Consider a Foucault pendulum of length \(15 \mathrm{~m}\) with a 110 -kg brass bob. It is set to swing with an amplitude of \(3.5^{\circ}\) a) What is the period of the pendulum? b) What is the maximum kinetic energy of the pendulum? c) What is the maximum speed of the pendulum?

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