A body oscillates with simple harmonic motion according to the following equation. x = (2.5 m) cos[(6π rad/s)t + π/4 rad] (a) At t = 7.0 s, find the displacement. 1.77 Correct: Your answer is correct. m (b) At t = 7.0 s, find the velocity. -33.32 Correct: Your answer is correct. m/s (c) At t = 7.0 s, find the acceleration. -628.1 Correct: Your answer is correct. m/s2 (d) At t = 7.0 s, find the phase of the motion. rad (e) At t = 7.0 s, find the frequency of the motion. Hz (f) At t = 7.0 s, find the period of the motion. s

Explanation:

Given:

Displacement, x = (2.5 m) cos[(6π rad/s)t + π/4 rad]

A.

At t = 7s,

x = 2.5 × cos(42π + π/4)

= 2.5 × cos(169/4 × π)

= 5/4 × sqrt2

= 1.77 m

B.

dx/dt = v = -(2.5 × 6π) × sin[(6π rad/s)t + π/4 rad]

= -15π × sin[(6π rad/s)t + π/4 rad]

At t = 7s,

= -15π × sin[(42π rad/s)t + π/4 rad]

= -15π × sin(169/4 × π)

= -15/2 × π × sqrt2

= -33.32 m/s

C.

dv/dt = a = -(2.5 × (6π)^2) × cos[(6π rad/s)t + π/4 rad]

= -90 × (π)^2) × cos[(6π rad/s)t + π/4 rad]

At t = 7s,

= -90 × (π)^2) × cos[(42π rad/s) + π/4 rad]

= -45 × (π)^2) × sqrt2

= -628.1 m/s^2

D.

Comparing ,

x = Acos(wt + phil)

With,

x = (2.5 m) cos[(6π rad/s)t + π/4 rad]

Phase angle, phil = π/4 rad

Since 2π rad = 360°

π/4 rad = 360/8

= 45°

E.

angular velocity, w = 2π/t

= 2π × f

Comparing the above equations,

w = 6π rad/s

Frequency, f = 6π/2π

= 3 Hz

F.

Period, t = 1/f

= 1/3

= 0.33 s.

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .

the answer would be 32*f