Question

My Notes The displacement from equilibrium of an oscillating weight suspended by a spring is given by y(t) = 1 4 cos(5t) where y is the displacement (in feet) and t is the time (in seconds). Find the displacement when t = 0, t = 1 4 , and t = 1 2 . (Round your answers to four decimal places.)
a) t = 0 y(0) = ft
(b) t = 1 4 y 1 4 = ft
(c) t = 1 2 y 1 2 = ft

Answer 1

Answer:

y(0) = 0.25 feet

$y(\frac{1}{4}) = 0.0789 \text{ feet}$

$y(\frac{1}{2}) =-0.2003 \text{ feet}$

Step-by-step explanation:

We are given the following information in the question:

The displacement from equilibrium of an oscillating weight suspended by a spring =

$y(t) = \displaystyle\frac{1}{4} \cos(5t)$

where y is the displacement in feet and t is the time in seconds.

Here, cos is in radians.

1) t = 0

$y(0) = \displaystyle\frac{1}{4} \cos(5(0)) = \frac{1}{4} \cos(0) = \frac{1}{4}(1) = \frac{1}{4}$

y(0) = 0.25 feet

2) t = $\frac{1}{4}$

$y(\displaystyle\frac{1}{4}) = \displaystyle\frac{1}{4} \cos(5(\frac{1}{4})) = \frac{1}{4} \cos(1.25) = \frac{1}{4}(0.31532236) =0.07883059$

$y(\frac{1}{4}) = 0.0789 \text{ feet}$

3) t = $\frac{1}{2}$

$y(\displaystyle\frac{1}{2}) = \displaystyle\frac{1}{4} \cos(5(\frac{1}{2})) = \frac{1}{4} \cos(2.5) = \frac{1}{4}(-0.80114362) = -0.200285905$

$y(\frac{1}{2}) =-0.2003 \text{ feet}$

The negative sign indicates the opposite direction of displacement.