Question

A force of 300 N is used to stretch a horizontal spring with a 0.5 kg block. The length of spring is 0.25 m. The block is released from rest and undergoes simple harmonic motion. Calculate the velocity when the block is stretched 15 cm.

Answer 1

Answer:

   v = 11.34 m / s

Explanation:

The simple harmonic motion of a spring and a mass is described by the equation

          x = A cos (wt + Фfi)

where A is the amplitude of movement in this case 0.25 m, w the angular velocity and fi the initial phase

the angular velocity is

           w = √ k / m

We can use Hooke's law to find the constant ka

           F = k x

where the force is 300N and the spring stretch is 0.25m

           k = F / x

           k = 300 / 0.25

           k = 1200 N / m

To find the phase angle di, let's use the system speed

         va = dx /dt

         va = A w sin (wt + Ф)

they tell us that the spring comes out of rest at time zero

         Vd = Aw sin Ф

the only way this term is zero is that the angle Ф = 0

substitutions in the first equations

          x = A cos wt

with

      w = √RA (1200 / 0.5)

      w = 48.99 rad / sec

we substitute in the first equations

        x = 0.25 cos (48.99 t)

speed is

         v = 0.25 48.99 without 48.99i

         

ask the speed for x = 0.15 m

we start by calculating the time it takes to get to this point

            x = A cos wt

            t = 1 / w cos-1 x / A

         we look for the time

           t = 1 / 48.99 cos-1 (0.15 / 0.25)

           t = 0.0189 s

this is the first time it takes to get to the requested point

    now we can calculate the speed

     v = Aw sin (wt)

          v = 0.25 48.99 sin (48.99 0.0189)

           v = 11.34 m / s