Simple harmonic motion of a spring mass

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SUMMARY

The amplitude of the simple harmonic motion for the spring-mass system is definitively 0.02m, as it represents the maximum displacement from the equilibrium position after the spring is pulled and released. The spring, initially 1m long, extends 0.1m under a 2kg load, but the amplitude is determined by the additional displacement of 0.02m. The discussion also clarifies the wave equation for a particle Q, which is positioned x length away from particle P, emphasizing that the phase difference is subtracted due to the wave's direction of travel in the Ox-positive direction.

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  • Understanding of Hooke's Law and spring mechanics
  • Familiarity with simple harmonic motion (SHM) concepts
  • Knowledge of wave equations and phase differences
  • Basic trigonometry and calculus for wave functions
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Kurokari
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Homework Statement



A spring of negligible mass have an original length of 1m. A load of 2kg is later added to it, resulting in an extension of 0.1m. Then it is pulled 0.02m and released.

What is the amplitude of the simple harmonic motion.

erm basically I just want to know which is the amplitude, 0.1m or 0.02m. Based on what I am told, the amplitude is 0.02m, but I just don't get it. It would be nice if someone can explain it =)

Homework Equations


x = x0 sin(ωt)

The Attempt at a Solution



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==================================================

The next question isn't really a homework question but an attempt on understanding the formula for a progressive wave.

Direction of wave is in the Ox-positive direction

For a particle, P, which is vibrating in a progressive wave with equation
y = asin(ωt) when t = 0, x = 0

The equation for Q particle, which is x length away from particle P with equation
y = asin(ωt - [(2∏x)/λ] ).

Question: Why does it minus the phase difference? Because to my understanding, if you want the equation of particle Q, you would add the phase different?

Again, your explanation and patience with me is greatly appreciated =)
 
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The amplitude is the extension beyond the equilibrium point. Hence it is 0.02m.
 
Kurokari said:
Direction of wave is in the Ox-positive direction
The equation for Q particle, which is x length away from particle P with equation
y = asin(ωt - [(2∏x)/λ] ).

Question: Why does it minus the phase difference? Because to my understanding, if you want the equation of particle Q, you would add the phase different?

Your equation for Q is for a particle 'x length away from particle P' and ON THE POSITIVE X SIDE OF of P.

Since the wave is traveling in the Ox-positive direction and Q is assumed to be further on the positve x-dir, then the time of oscillation for Q is less than that for P.

Hence the shm for Q is given by the same as that for P but for a shorter time (t - t') where t' is the time the wave takes to travel from P to Q i.e. to travel 'length x'.

This gives the shm for Q to be

y = asin[ω(t - t')].

One or two further steps one gets the equation given for Q.
 

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