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just olya [345]
2 years ago
5

A 300 g wooden block on a smooth, level surface is firmly attached to a very light horizontal spring with a spring constant of 2

00 N/m.
The spring-block system is initially stretched a distance “d” beyond the spring's equilibrium position and is released from rest. As the block moves to the right, its speed is 50.0 cm/s when the spring-block system is 4.00 cm from the equilibrium position.
Find d, the initial displacement of the spring-block system (in cm).

Physics
1 answer:
Tom [10]2 years ago
4 0

the solution for the oscillatory movement allows to find the result for the amplitude of the initial displacement is:

  • The range of motion is: A = 4.44 cm

<h3>Oscillatory movement.</h3>

The oscillatory periodic motion of a system occurs when there is a recovered force, in the special case that this force is proportional to the displacement is called simple harmonic motion, which is described by the expression.

            x = A cos (wt + Ф)

            w² = k/m

where x is the displacement, A the amplitude, w the angular velocity, t the time, k the spring constant, m the mass, and Ф a phase constant determined by the initial conditions.

Let's find the angular velocity/

            w= \sqrt{ \frac{200}{0.300} }

            w = 25.8 rad/s

Let's look for the constant Ф, as the system is released from rest its initial velocity is zero, for zero time. The definition of speed is:

             v= \frac{dx}{dt}

             v= - A w sin (wt +Ф)

             

             0 = -A w sin Ф

            Ф= 0

They indicate that at a given instant of the time the velocity is v= 50.0 cm/s and it is in a position x= 4.00 cm, let us write the equations for this time

Position.

               4.00 = A cos 25.8t

Speed.

              50.0 = - At 25.8 sin 25.8t

To solve the system, ;et's square and add.

              Cos² 25.8t = \frac{16}{A^2}

              sin² 25.8t = \frac{3.756}{A^2 }

              1 = \frac{1}{A^2} \ (16 + 3.756)

               A = \sqrt{19.756}

               A= 4.44 cm

In conclusion using the solution for the oscillatory movement we can find the result for the amplitude of the initial displacement is:

The range of motion is: A = 444 cm

Learn more about oscillatory motion here:  brainly.com/question/14311816

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