Simple Harmonic Motion of an oscillating particle

[maniacp08]

The period of an oscillating particle is 56 s, and its amplitude is 18 cm. At t = 0, it is at its equilibrium position. Find the distances traveled during these intervals.
(a) t = 0 to t = 14 s
cm

(b) t = 14 s to t = 28 s
cm

(c) t = 0 to t = 7 s
cm

(d) t = 7 s to t = 14 s
cm

x = Acos($$\omega$$ * t * $$\delta$$)
$$\omega$$ = 2$$\pi$$ / T
$$\omega$$ = 2$$\pi$$ / 56

Since x = 0 when t = 0
Then $$\delta$$ = $$\pi$$ / 2
correct?

for part A)

x = 18cm * cos[2$$\pi$$ / 56 * 14s + $$\pi$$ / 2] - cos[2$$\pi$$ / 56 * 0s + $$\pi$$ / 2]

Is this good?

 

[alphysicist]

Hi maniacp08,

The period of an oscillating particle is 56 s, and its amplitude is 18 cm. At t = 0, it is at its equilibrium position. Find the distances traveled during these intervals.
(a) t = 0 to t = 14 s
cm

(b) t = 14 s to t = 28 s
cm

(c) t = 0 to t = 7 s
cm

(d) t = 7 s to t = 14 s
cm

x = Acos($$\omega$$ * t * $$\delta$$)
$$\omega$$ = 2$$\pi$$ / T
$$\omega$$ = 2$$\pi$$ / 56

Since x = 0 when t = 0
Then $$\delta$$ = $$\pi$$ / 2
correct?

for part A)

x = 18cm * cos[2$$\pi$$ / 56 * 14s + $$\pi$$ / 2] - cos[2$$\pi$$ / 56 * 0s + $$\pi$$ / 2]

The 18cm needs to multiply both cosine functions, not just the first one. Was that just a typo?

Also you can think about how the time period of 14 seconds is related to the period, to get the answer right away.