Question

A spring is attached to the ceiling and pulled 8 cm down from equilibrium and released. The amplitude decreases by 17% each second. The spring oscillates 19 times each second. Find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.

Answer 1

Answer:

D(t) = 8(0.83)^(t) cos 38πt

Explanation:

We are told that the spring oscillates 19 times each second.

Thus, period = 1/19

We are also told that it's pulled 8cm downwards and the amplitude decreases by 17% each second.

Thus;

Amplitude;A = 8 × (1 - (17/100))^(t)

A = 8(0.83)^(t)

If we consider the function;

y = A cos (bx - c) + d

Now, 2π/b = period

So, 1/19 = 2π/b

b = 38π

So, D(t) = 8(0.83)^(t) cos (38πt - c) + d

Since we started from minimum,

Vertical shift, d = 0 and horizontal shift c = 0

So,we now have;

D(t) = 8(0.83)^(t) cos 38πt