A complete description of simple harmonic motion must take into account several physical quantities and various mathematical relations among them. This information is needed to solve oscillation problems of this type.

The position of a 45 g oscillating mass is given by x(t)=(2.0cm)cos(10t), where t is in seconds.

a. Determine the velocity at t=0.40s.
b. Assume that the oscillating mass described in Part A is attached to a spring. What would the spring constant k of this spring be?
c. What is the total energy E of the mass described in the previous parts?

a) v = - 0.698 m/s, b) k= 45 N/m, c) E = 0.009J

Explanation: From the question, we have the mass as 45g = 0.45kg

The position x(t) is given as

x(t) = 2.0 cos (10t).

Let us compare the equation with the general equation of a periodic motion, we have that

x(t) = A cos (ωt)

x(t) = 2.0 cos (10t)

A = amplitude = 2.0cm = 0.02 m

(ωt) = (10t)

Hence, ω = 10 rad/s

A)

To get the velocity, we find the first derivative of displacement x(t).

Hence dx(t) /dt = - 20 sin (10t) {chain rule was used in differentiating}

So therefore, v = - 20 sin (10t), at t =0.40s, we have v as

v = - 20 sin (10×0.40)

v = - 20 sin 4

v = - 0.698 m/s

The negative sign denotes that the velocity is in opposite direction to the displacement.

B)

The formulae that relates angular frequency, spring constant and mass of a loaded spring is given below as

ω = √k/m

Where k = spring constant, m = mass of object = 0.45kg, ω = 10 rad/s

By substituting the parameters

10 = √k/0.45

By squaring both sides

10² = k/ 0.45

100 = k/0.45

k = 100 × 0.45

k = 45 N/m

C)

The total energy of a loaded spring is given below as

E = 1/2 KA²

where A = amplitude = 0.02m

E = 1/2 ×45 × 0.02²

E = 0.018/2

E = 0.009 J

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2