Trying to solve a harmonic motion problem

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SUMMARY

This discussion focuses on solving a harmonic motion problem involving a mass oscillating on a spring with given parameters: amplitude of 0.100 m, spring constant (k) of 8.00 N/m, and mass of 0.5 kg. The initial attempt to solve for the time elapsed for the mass to travel from x=0.00 to x=0.080 m using the cosine function yielded an incorrect result of 0.554 seconds. The correct solution, derived from the sine function, is 0.232 seconds, as confirmed by both the textbook and graph tracing. The discussion emphasizes the importance of recognizing multiple valid solutions when using inverse trigonometric functions in physics.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Familiarity with trigonometric functions and their inverses
  • Knowledge of oscillatory motion equations
  • Basic physics concepts related to mass-spring systems
NEXT STEPS
  • Study the derivation and application of the equations for simple harmonic motion
  • Learn about the properties and graphs of sine and cosine functions
  • Explore the implications of multiple solutions in inverse trigonometric functions
  • Investigate the velocity equations in harmonic motion to validate solutions
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to clarify concepts of harmonic motion and trigonometric applications in physics.

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"simple" harmonic motion

Trying to solve a harmonic motion problem (mass oscillating on a spring), given:
amplitude=.100 m
k=8.00 N/m
mass=.5 kg
to solve for time elapsed for the mass to travel from x=0.00 to x=.080 m.
It's easy to solve for w (omega) = sqrt(8/.5) = 4, and using the equation for shm from my physics textbook I got this equation:
.08 = .1 cos(4t-pi/2)
and then
t = ((arccos 0.8) + pi/2)/4 = .554 sec. which unfortunately is wrong. This turns out to be the time it takes for the mass to go past .08 m, past its maximum displacement of .1 m, and BACK to .08 m.

Now, this equation is equivalent:
.08 = .1 sin(4t)
Check it...the graphs are exactly the same. But this one solves to
t = .232 sec. which is the answer given by the book, and tracing the graph confirms that this is the correct answer.

So, there must be a way to solve the first equation [.08 = .1 cos(4t-pi/2)] to get the correct answer of .232 sec., but I can't see it.

What am I missing?
 
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When using inverse trig operations, be aware that there is more than one answer, for example: -

sinx = 0
x = arcsin(0)
x = n(pi) where n is an integer.

When applying such maths to physics, it is necessary to choose which answer you require, which may, or may not be obvious.

In your question, you assumed that the only solution to arccos(0.8) is 0.6435... when in fact, -0.6435 is an equally valid solution. (This is a consequence of cos(x) = cos(-x)). Substituting this into your equation gives the answer provided by the book.

If you are uncomfortable with having to choose which answer is correct without using educated guesswork or drawing graphs, use the equation for the velocity of the object to find out whether each solution corresponds to a positive or negative velocity.
 
Thanks Claude, that was very helpful.
 

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