Difficulty computing second derivative value in SHM problem

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Homework Help Overview

The problem involves simple harmonic motion, specifically analyzing the displacement function x(t) = 5cos(30t) + 4sin(30t) and computing its second derivative. The original poster is attempting to find the t-values where the second derivative x''(t) equals zero.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster expresses frustration over obtaining an incorrect sign for a t-value when setting the second derivative to zero, despite having successfully completed similar steps with the first derivative. Some participants question the clarity of the original poster's intent regarding finding t-values versus x-values, while others suggest verifying the solutions derived from squaring equations.

Discussion Status

The discussion is ongoing, with the original poster clarifying their intent after initial confusion. Participants are providing guidance on potential pitfalls in the algebraic process, particularly regarding the implications of squaring equations and ensuring all solutions are accounted for.

Contextual Notes

The original poster notes a sense of urgency due to the late hour and expresses uncertainty about their algebraic work, indicating a possible constraint in their ability to review their calculations thoroughly.

JDC123
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Homework Statement



The displacement of a machine is given by the simple harmonic motion as x(t) = 5cos(30t)+4sin(30t). Find the amplitude of motion, and the amplitude of the velocity.

Homework Equations



x''(t) = -4500cos(30t)-3600sin(30t)

The Attempt at a Solution


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I should note that I'm not having any difficulty with the problem conceptually, as it is very simple. However, for some reason unknown to me, I'm getting the wrong sign on a t-value when I set x''(t) to zero. I've checked and rechecked my work numerous times, and I'm not sure what's going on. This exact same procedure worked with the first derivative, so I don't get why it's failing me with the second. I've graphed x''(t) on Desmos, and the first positive t-value for x''(t)=0 is 0.0749. My answer is equal to 0.02986. Strangely, according to the graph, the first negative t-value to satisfy x''(t)=0 is -0.02986. Does anyone know why I'm getting the wrong sign? It's driving me absolutely insane. Please see the attached photo to view my attempt at a solution.

It is admittedly quite late here, and while there must be some pernicious error in my algebra, I can't find it.Thank you!
 

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It is unclear what you mean when you say that you want to find the x-values for which x'' is zero. This should not happen for negative or for positive x. Try writing x" as a function of x instead of t.
 
My apologies! I'm after t-values for x''(t)=0, not x-values. I have edited my original post to correct this mistake.
 
Last edited:
When you square an equation you must check that the solutions satisfy the original equation. Later you are missing one of the solutions when taking the square root and the solution you are left with happens to be the false root you introduced by squaring. I would quote the specific point in your work but this is impossible since you just posted an image. Generally, note that only posting photos of your work breaks the PF guidelines for homework and that such posts may be deleted.
 

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