An object attached to a coiled spring is pulled down 5 centimeters from its rest position and released. If the motion is simple harmonic in nature, with a period of seconds

Question

An object attached to a coiled spring is pulled down 5 centimeters from its rest position and released. If the motion is simple harmonic in nature, with a period of seconds, answer the following questions:

a. What is the maximum displacement form equilibrium of the object?

b. What is the time required for one oscillation?

c. What is the frequency?

d. Write an equation to model the motion of the object.

Answer 1

max displacement = 5 centimeters

period = pi = about 3.141593... seconds = about 3 seconds

frequency = inverse = 1/pi = about 0.31831 hz

equation that fits is a GP with r = common ratio

an = a1(r^n -1)/(r-1) with a1 = 1/2, an = the length of the nth oscilation which approaches zero as n gets ever larger

but the problem says simple harmonic motion which contradicts the other info

so use a sinusoidal function h(t) = Asin(2pit/B + C) + D where A = Amplititude = (max -min)/2

D = midline = (max+min)/2 = vertical shift, B = period, C= horizontal shift

this also contradicts basic physics as it becomes a perpetual motion machine

Answer 2

The maximum displacement is the same as the initial displacement (5 cm). This is the "A" value or amplitude of the simple harmonic motion.

We are given the period and the period is the time required to complete one oscillation so that is π seconds (approx 3.14 seconds).

The frequency is 1/period therefore it is 1/π hz (approx 0.318 hz).

To write an equation that models the motion, you must set up a coordinate plane. I'll set it up like this:

Therefore, when initially stretched the position is -5cm. This fits a "negative cosine" function very nicely so:

y = -5cos(2t). Note that the amplitude goes outside the cosine function as a multiplier and that the multiplier of the "t" variable is 2π/period and 2π/π = 2