The 0.18 kg block oscillates back and forth along a straight line on a frictionless horizontal surface. Its displacement from the origin is given by the following equation. x = (11 cm)cos[(15 rad/s)t + π/2 rad)] (a) What is the oscillation frequency? 2.39 Correct: Your answer is correct. Hz (b) What is the maximum speed acquired by the block? 165 Correct: Your answer is correct. cm/s At what value of x does this occur? cm (c) What is the maximum acceleration of the block? cm/s2 At what values of x does this occur? (positive then negative) cm and cm (d) What force, applied to the block, results in the given oscillation? ( N/m)x

(a) f = 2.39 Hz

(b) Vmax = 165 cm/s

(c) amax = 24.75 m/s²

Explanation:

Given:

Mass, m = 0.18kg

displacement from the origin = ( 11 cm) cos [ ( 15 rad/s) t + (π / 2 rad)          -1

(a)

15 rad/s  will be the angular frequency, ω

So frequency is

f = ω / 2π

f = 15 / 2π

f = 2.39 Hz

(b)

we know,

Vmax = ω xmax

From equation 1, the value of xmax is 11cm

xmax = 0.11m

Vmax = 15 rad/s X 0.11m

Vmax = 1.65 m/s

Vmax = 165 cm/s

(c)

Maximum acceleration, amax = ω²xm

amax = (15)² X 0.11m

amax = 24.75 m/s²

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .