High School Differentiating some simple harmonic equation

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SUMMARY

The discussion centers on differentiating the simple harmonic motion equation x = x0sin(wt), where w represents angular frequency and x0 denotes maximum displacement. The user initially derived the acceleration as a = -w^2x0sin(wt) but later realized the correct expression should be a = -w^2x. This highlights the importance of careful differentiation in physics equations, particularly in the context of simple harmonic motion.

PREREQUISITES
  • Understanding of simple harmonic motion principles
  • Familiarity with differentiation in calculus
  • Knowledge of angular frequency (w) and maximum displacement (x0)
  • Ability to manipulate trigonometric functions
NEXT STEPS
  • Study the principles of simple harmonic motion in detail
  • Practice differentiation techniques for trigonometric functions
  • Explore the relationship between acceleration and displacement in harmonic motion
  • Learn about the implications of angular frequency in oscillatory systems
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in mastering calculus applications in oscillatory motion.

Trance-
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So I was just trying to differentiate (for no good reason) the equation :
x=x0sin(wt)
(w= angular frequency, x0= maximum displacement, t=time)

to obtain the expression :
a= -w2x

I differentiated twice with respect to time the initial expression for x and got:
a= -w2x0sin(wt)

I must have done something wrong while differentiating... so help me out, folks.
 
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I'll write it in a more suggestive manner:
$$a = -\omega^2 (x_0 \sin(\omega t))$$
Notice anything?
 
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axmls said:
I'll write it in a more suggestive manner:
$$a = -\omega^2 (x_0 \sin(\omega t))$$
Notice anything?

Oh crap!
My sleep deprivation must be taking its toll dammit. Thanks a lot man!
I make the dumbest mistakes.
 
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Relativistic Momentum, Mass, and Energy Momentum and mass (...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...) The momentum of a particle moving with velocity ##v## is given by $$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$ ENERGY In relativistic mechanics, as in classic mechanics, the net force on a particle is equal to the time rate of change of the momentum of the particle. Considering one-dimensional...

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