A particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the origin, at t = 0 and moves to the right. The amplitude of its motion is 2.60 cm, and the frequency is 1.30 Hz.
(f) Determine the total distance traveled between t = 0 and t = 1.15 s.
PLEASE HELP!

Question

A particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the origin, at t = 0 and moves to the right. The amplitude of its motion is 2.60 cm, and the frequency is 1.30 Hz.
  (f) Determine the total distance traveled between t = 0 and t = 1.15 s.
PLEASE HELP!

mos1635 · Answer

x=Asinωt
A=2.6 cm
ω=2*π*f=2*3.14*1.3=8.164 rad/sec
at t=1.15 sec partical will be at position 
x=2.6 sin 8.164*1.15
x=2.6 sin10.79689
x=2.6 * 0.18732799
x=0.487 cm
now from f and t we calculate N=number of full oscilations
f=N/t
N=f*t
N=1.3*1.15
N=1.495
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total distance= 4A+A+(A-x)=6A-x=6*2.6-0.487=15.113cm