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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.

Express your answer in meters per second to two significant figures.

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?

Express your answer in newtons per meter to two significant figures.

C

What is the total energy E of the mass described in the previous parts?

Express your answer in joules to two significant figures.

Respuesta :

Answer:

A)  v = 15 cm / s

, B) k = 4.5 N / m, C)  Em = 9.0 10⁻⁴ J

Explanation:

A) The position of the harmonic oscillation given in the data by the equation

      x = 2 cos (10t)

The speed is given by

        v = dx / dt

        v = 2 10 (- sin 10t)

        v = -20 sin 10t

We calculate for the time 0.4 s

        v = -20 sin (10 0.40)

Remember that the angles are in radians

       v = -20 (-0.7568)

       v = 15.14 cm / s

       v = 15 cm / s

b) Of the equation the angular velocity is

       w = 10 rad / s

It is related to physical quantities

     w = RA k / m

     k = m w2

     k = 45 10-3 10 2

    k = 4.5 N / m

c) Mechanical energy is

      Em = ½ k A²

     

      A = 2.0 cm = 0.02 m

     Em = ½ 4.5 0.02²

     Em = 9.0 10⁻⁴ J

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