Simple Harmonic Motion and phase constant

[Dalip Saini]

A simple harmonic oscillator consists of a block of mass 45 g attached to a spring of spring constant 240 N/m, oscillating on a frictionless surface. If the block is displaced 3.5 cm from its equilibrium position and released so that its initial velocity is zero, what is the phase constant, φ , of its oscillations? Assume the block moves according to the equation x=xmaxcos(ωt+φ) .

• A :
  
  45o
• B :
  
  30o
• C :
  
  0o
• D :
  
  90o
• E :
  
  The phase constant is arbitrary.

 

[jbriggs444]

What do you think?

 

[Dalip Saini]

Im really stuck on how to solve this problem. I know that the Xm is 0.035m. I also know that the w can be solved for using sqroot(k/m). But I'm stuck now on how to use this information to solve for the answer

 

[jbriggs444]

You should not need to do any arithmetic to answer this problem. You are not asked for omega.

Assume that "initial" means t = 0. What is the situation at t=0?

 

[Dalip Saini]

since the initial velocity is zero, would u take the derivative of the function to get v=-wxmsin(wt+φ). Then plug in the values to get 0 = -240(0.035)sin[(240)(0)+φ). Then solve for φ and get zero. According to the assignment the correct answer is zero. Is this the correct way to solve it?

 

[jbriggs444]

since the initial velocity is zero, would u take the derivative of the function to get v=-wxmsin(wt+φ). Then plug in the values to get 0 = -240(0.035)sin[(240)(0)+φ). Then solve for φ and get zero. According to the assignment the correct answer is zero. Is this the correct way to solve it?

Yes, that is one correct way.

Another approach would be to convince yourself that the starting position is a position of maximum displacement and that the cosine function is maximized when its argument is zero.

 

[Dalip Saini]

ok thank you