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A student is observing the oscillations of a mass on the end of a spring. The spring has a force constant of 6 x 10-5N/m, and the mass is 0.15 kg. She pulls it 6 cm below equilibrium and lets it go. She now writes an equation which she believes describes the motion of the mass. In her description, the y axis is vertical, and UP is positive. Her equation is y = (3 cm) sin ( 0.02 sec-1 t + 3.14) (A) Does she have the amplitude right? If not, what was right? If yes, explain. (B) Does she have the frequency right? If not, what was right? If yes, explain. (C) Does she have the phase angle right? If not, what was right? If yes, explain.

Answer :

Answer:

a)  A = 6 cm, b)  w  = 0.02 s⁻¹, c)  φ = 0º

The correct equation is     y = 6 cos (0.02 t)

Explanation:

a) The range of motion is the amount the body is pulled, as the body is pulled 6 cm

         A = 6 cm

The value written in the equation is not correct

       

b) the angular velocity is given by

              w = √ k / m

              w = √ 6 10⁻⁵ / 0.15

              w = 2 10⁻² rad / s = 0.02 s⁻¹¹

The frequency is  correct

c) the phase constant can be found from the initial conditions, at time t = 0 the position of the body corresponds to the amplitude

             x = A cos (wt + φ)

             t = 0 s

             A = A cos φ

             Cos φ = 1

             φ = 0º

The result is not correct

The correct equation is

              Y = 6 cos (0.02 t)

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