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.

 

[thomate1]

Yes, your approach is correct. Here is a breakdown of the steps for each part:

(a) t = 0 to t = 14 s
To find the distance traveled during this interval, we can use the formula x = Acos(\omega * t), where A is the amplitude and \omega is the angular frequency. Plugging in the given values, we have x = 18cm * cos[(2\pi / 56) * 14s]. This simplifies to x = 18cm * cos(0.5\pi) = 18cm * 0 = 0cm. Therefore, the particle does not travel any distance during this interval as it remains at its equilibrium position.

(b) t = 14 s to t = 28 s
Using the same formula, we have x = 18cm * cos[(2\pi / 56) * 28s]. This simplifies to x = 18cm * cos(\pi) = 18cm * (-1) = -18cm. Therefore, the particle travels a distance of 18cm to the left during this interval.

(c) t = 0 to t = 7 s
Using the formula x = Acos(\omega * t), we have x = 18cm * cos[(2\pi / 56) * 7s]. This simplifies to x = 18cm * cos(0.25\pi) = 18cm * 0.7071 = 12.73cm. Therefore, the particle travels a distance of 12.73cm to the right during this interval.

(d) t = 7 s to t = 14 s
Using the same formula, we have x = 18cm * cos[(2\pi / 56) * 14s] - 18cm * cos[(2\pi / 56) * 7s]. This simplifies to x = 18cm * cos(0.5\pi) - 18cm * cos(0.25\pi) = 0cm - 12.73cm = -12.73cm. Therefore, the particle travels a distance of 12.73cm to the left during this interval.