Simple Harmonic Motion

  • Thread starter shan
  • Start date
  • #1
shan
57
0
I just need someone to tell me if I've done the problem right since the answer is mysteriously missing

The question is about a rotating wheel causing a piston to execute SHM inside a cylinder (the piston being connected to the outside of the wheel). The wheel rotates with an angular speed of 12rads-1 and it has a radius of 30cm.

What will be the distance of the piston from an end position after 0.10s?

ω = θ/t
θ = 12x0.1 = 1.2rad

displacement from equilibrium point = amplitudexCosθ = 0.3xCos1.2 =0.1087m

so distance from end position = 0.3-0.1087 = 0.19m (2sf)
 

Answers and Replies

  • #2
lavalamp
279
1
Well I've read through it and it looks OK to me. Couldn't you just have asked your teacher though?
 
  • #3
shan
57
0
School's out and I'm on my study leave...
 
  • #4
lavalamp
279
1
Lucky you, I'm in school until almost Christmas. Then I have exams (with very little study leave) in January.
 

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