Question

P(t)P(t)P, left parenthesis, t, right parenthesis models the distance of a swinging pendulum (in \text{cm}cmstart text, c, m, end text) from the place it was released ttt seconds after it starts to swing. Here, ttt is entered in radians. P(t) = -5\cos\left(2\pi t\right) + 5P(t)=−5cos(2πt)+5P, left parenthesis, t, right parenthesis, equals, minus, 5, cosine, left parenthesis, 2, pi, t, right parenthesis, plus, 5 What is the first time the pendulum reaches 3.5\text{ cm}3.5 cm3, point, 5, start text, space, c, m, end text from the place it was released? Round your final answer to the nearest hundredth of a second.

Answer 1

0.20second

Explanation:

Given the distance of a swinging pendulum modeled by the equation P(t)=−5cos(2πt)+5, to get the first time the pendulum reaches 3.5cm, we will have to substitute P(t)= 3.5cm into the given equation and calculate the value of 't' as shown;

3.5 = −5cos(2πt)+5

Subtracting 5 from both sides;

3.5-5 = −5cos(2πt)+5-5

-1.5 = −5cos(2πt)

Dividing both sides by -5 will give;

`(-1.5)/(-5) = cos2\pi t\\0.3 = cos2\pi t\\2\pi t=cos^(-1) 0.3\\2\pi t = 72.54^(0) \\360t = 72.54^(0) \\t = (72.54)/(360) \\t = 0.20secs`

The first time the pendulum reaches 3.5cm is after 0.20secs