A solid metal block with a mass of 1.30 kg is attached to a spring and is able to oscillate horizontally with negligible friction. The block is pulled to a distance of 0.200 m from its equilibrium position, held in place with a force of 19.0 N, and then released from rest. It then oscillates in simple harmonic motion. (The block oscillates along the x-axis, where x = 0 (e) What is the maximum acceleration of the block? (Enter the magnitude in m/s2.) m/s2 (f) At what position(s) (in m) on the x-axis does the maximum acceleration occur? (g) What is the total mechanical energy of the oscillating spring-block system (inJ)? h) What is the speed of the block (in m/s) when its position is equal to one-third of the maximum displacement from equilibrium? m/s () What is the magnitude of the acceleration of the block (in m/s) when its position is equal to one-third of the maximum displacement from equilibrium? m/s2 is the equilibrium position.)

e)   a = 14.62 m / s², f)  the maximum acceleration occurs at the points x = ± A , g)    Em = 1.9 J , h)   v = 1.61 m / s, i) a = -4.82 m / s²

Explanation:

A system of a spring with a mass is described by the equation

             x = A cos (wt + Ф)

Where

           w = √ k / m

e) ask to find the acceleration, for this we use the definition of acceleration

            v = dx / dt

            a = dv / dt

            v = -A w cos (wt + Ф)

            a = -A w² cos (wt + Ф)

The acceleration is maximum when the cosine is ± 1

            a = A w²

In the exercise, indicate that the amplitude is A = 0.200 m

To find the constant k, let's use Newton's equilibrium equation

          F - Fe = 0

          F = k x

          k = F / x

          k = 19 / 0.2

          k = 95 N / m

Now we can calculate the angular velocity

          w = √ k / m

         w =√ (95 / 1.3)

         w = 8.55 rad / s

Substitutes in the maximum acceleration equation

          a = 0.2 8.55²

          a = 14.62 m / s²

f) the acceleration depends on the cosine and the cosine is maximum in the express of the displacement, therefore the maximum acceleration occurs at the points x = ± A

g) mechanical energy is given by the equation

         Em = ½ k A²

         Em = ½ 95 0.2²

         Em = 1.9 J

h) what is the speed when the position is x = 1/3 A

        Em = K + U

        Em = ½ m v² + ½ k x²

        v² = (Em - ½ k x²) 2 / m

        v² = (1.9 - ½ 95 (0.2 /3)²)  2 / 1.30

        v² = 2,598

        v = 1.61 m / s

i) block acceleration when x = A / 3

For this we must find fi and the time to reach this position

Let's look fi

As the system starts from rest the speed starts is zero

         V = A w sin (0+ Ф)

          0 = sin Ф

         Ф = 0

Now we look for the time to reach the displacement x = A / 3

           A / 3 = A cos 8.5 t

           1/3 = cos 8.5t

           8.5t = cos-1 1/3

           t = 1 / 8.5 cos-1 1/3

           t = 0.1448 s

Now we substitute in the acceleration equation

          a = - A w² cos wt

          a = - 0.2 8.5² cos (8.5 0.1448)

          a = -4.82 m / s²

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.