    # 3. At t = 0, an 885-g mass at rest on the end of a horizontal spring (k = 184 N/m) is struck by a hammer which gives it an initial speed of 2.26 m/s. Determine (a) the period and frequency of the motion, (b) the amplitude, (c) the maximum acceleration, (d) the total energy, and (e) the kinetic energy when x = 0.4A where A is the amplitude.

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4 months ago  Answer:(a) Period = 0.436 s, frequency = 2.29 Hz(b) 0.157 m(c) 32.6 m/s²(d) 2.26 J(e) 1.91 JExplanation:This is s simple harmonic motion in the form of a loaded spring. The initial velocity is the maximum velocity because the maximum velocity occurs at the equilibrium point.Given:Mass, m = 885 g = 0.885 kgSpring constant, k = 184 N/mMaximum velocity, = 2.26 m/s(a) Period, T, is given by:  Frequency, f, is given by (b) The maximum velocity is related to the amplitude by where A is the amplitude ω is the angular velocity and is given by  (c) Maximum acceleration is given by  (d) Because the total energy is equal at any time, we determine the energy at equilibrium.At equilibrium, displacement is 0 m and velocity is maximum. Hence, the potential energy is 0 J and the kinetic is maximum which is determined thus: The maximum energy is 2.26 J.(e) The velocity at any point is 0 At point x = 4A, 1 Putting the values of A and ω, 2 3 The kinetic energy is 4                1 month ago 10 5.0     