1. The problem statement, all variables and given/known data Through simple harmonic motion produced on a trampoline Ivan ossilates freely, with a period of 0.85s. On a reference circle show Ivans position after 1/8th of a cycle. Label (and solve) the distance of this position from the equilibrium position.(Max displacement is 0.14m). C and A are labelled as the maximum displacement, and B the equilibrium position 2. Relevant equations 3. The attempt at a solution I drew a reference circle, creating a right-angled triangle with an angle to the vertical of 45 degrees. The answer does not make much sense to me. Ill submit it as an attachment.
When one uses a reference circle to derive the SHM equations it is assumed that the radius vector rotates at a constant angular velocity [tex]\omega[/tex]. Usually one takes it to rotate anticlockwise from some reference line. If the reference line is the horizontal line running throught the diameter of the reference circle one would start the time measurement when the radius vector passes the three o'clock position. This would then give the formula [tex]y = a \sin(\theta)[/tex] for the displacement where [tex]theta[/tex] is the angle that the radius vector is making with the reference line. If one chooses to start the time measurement when the radius vector is at the twelve o'clock position the formula would use the cosine rather than the sine of the angle (which is then measured from the y-axis). In this case a is the amplitude of the motion which is 0.14 m. For a complete cycle we have that [tex]\theta = 2 \pi[/tex] radians which means that for one eight of a cycle the angle would be [tex]\theta = \frac{\pi}{4}[/tex] radians which is [tex]45^o[/tex] which means that the displacement would be [tex]y = 0.14\sin(\frac{\pi}{4})[/tex] meters after one eight of a cycle.
Important: should'n forget to position your calculator mode in radians, since equation deals with angular displacement theta in radians.