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Chapter 10: Problem 34

Show that, for SHM, the maximum displacement, velocity, and acceleration are related by \(v_{\mathrm{m}}^{2}=a_{\mathrm{m}} A\).

Short Answer

Expert verified
Question: Prove the relationship \(v_{\mathrm{m}}^{2}=a_{\mathrm{m}} A\) for Simple Harmonic Motion by analyzing the maximum displacement (A), maximum velocity (v_m), and maximum acceleration (a_m). Answer: By analyzing the equations for displacement, velocity, and acceleration in Simple Harmonic Motion and finding their maximum values, we proved that the relationship \(v_{\mathrm{m}}^{2}=a_{\mathrm{m}} A\) holds true.

Step by step solution

01

Write down the equations for displacement, velocity, and acceleration in SHM

We'll start by writing down the equations for displacement (x), velocity (v), and acceleration (a) in SHM. Displacement: \(x(t) = A \cos(\omega t + \phi)\) Velocity: \(v(t) = -A\omega \sin(\omega t + \phi)\) Acceleration: \(a(t) = -A\omega^2 \cos(\omega t + \phi)\) Here, A represents the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase.
02

Calculate the maximum velocity

In order to find the maximum velocity (v_m), we need to find the maximum value of the velocity function \(v(t)\). This would be when the sine function is equal to its maximum value (±1). Therefore, \(v_{\mathrm{m}}=\pm A\omega\)
03

Calculate the maximum acceleration

Similar to Step 2, in order to find the maximum acceleration (a_m), we need to find the maximum value of the acceleration function \(a(t)\). This would be when the cosine function is equal to its maximum value (±1). Therefore, \(a_{\mathrm{m}}=\pm A\omega^{2}\)
04

Show that \(v_{\mathrm{m}}^{2}=a_{\mathrm{m}} A\)

Now, we will substitute the expressions for \(v_{\mathrm{m}}\) and \(a_{\mathrm{m}}\) we calculated in Steps 2 and 3 into the given equation. \(v_{\mathrm{m}}^{2} = (\pm A\omega)^{2} = a_{\mathrm{m}} A = (\pm A\omega^{2})A\) Simplify and compare: \((\pm A\omega)^{2} = (\pm A\omega^{2})A\) \((\pm A^2\omega^2) = (\pm A^2\omega^2)\) Hence, the given relationship \(v_{\mathrm{m}}^{2}=a_{\mathrm{m}} A\) is proved for Simple Harmonic Motion.

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Most popular questions from this chapter

A pendulum clock has a period of 0.650 s on Earth. It is taken to another planet and found to have a period of \(0.862 \mathrm{s}\). The change in the pendulum's length is negligible. (a) Is the gravitational field strength on the other planet greater than or less than that on Earth? (b) Find the gravitational field strength on the other planet.
A \(0.50-\mathrm{kg}\) mass is suspended from a string, forming a pendulum. The period of this pendulum is 1.5 s when the amplitude is \(1.0 \mathrm{cm} .\) The mass of the pendulum is now reduced to \(0.25 \mathrm{kg} .\) What is the period of oscillation now, when the amplitude is \(2.0 \mathrm{cm} ?\) (W tutorial: change in period)
A 91 -kg man's thighbone has a relaxed length of \(0.50 \mathrm{m},\) a cross- sectional area of \(7.0 \times 10^{-4} \mathrm{m}^{2},\) and a Young's modulus of \(1.1 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) By how much does the thighbone compress when the man is standing on both feet?
The diaphragm of a speaker has a mass of \(50.0 \mathrm{g}\) and responds to a signal of frequency \(2.0 \mathrm{kHz}\) by moving back and forth with an amplitude of \(1.8 \times 10^{-4} \mathrm{m}\) at that frequency. (a) What is the maximum force acting on the diaphragm? (b) What is the mechanical energy of the diaphragm?
The period of oscillation of an object in an ideal spring-and-mass system is \(0.50 \mathrm{s}\) and the amplitude is \(5.0 \mathrm{cm} .\) What is the speed at the equilibrium point?
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