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Simple harmonic motion of a spring mass 
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#1
Oct1711, 10:39 AM

P: 35

1. The problem statement, all variables and given/known data
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 =) 2. Relevant equations x = x_{0} sin(ωt) 3. The attempt at a solution  ================================================== 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 Oxpositive 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 =) 


#2
Oct1711, 10:50 AM

P: 950

The amplitude is the extension beyond the equilibrium point. Hence it is 0.02m.



#3
Oct1711, 11:04 AM

P: 950

Since the wave is travelling in the Oxpositive direction and Q is assumed to be further on the positve xdir, 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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