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 J
Explanation:
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 kg
Spring constant, k = 184 N/m
Maximum velocity, [tex]v_m[/tex] = 2.26 m/s
(a) Period, T, is given by:
[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]
[tex]T=2\pi\sqrt{\dfrac{0.885 \text{ kg}}{184\text{ N/m}}} = 0.436 \text{ s}[/tex]
Frequency, f, is given by
[tex]f= \dfrac{1}{T} = \dfrac{1}{0.436\text{ s}} = 2.29\text{ Hz}[/tex]
(b) The maximum velocity is related to the amplitude by
[tex]v_m=\omega A[/tex]
where A is the amplitude ω is the angular velocity and is given by
[tex]\omega=2\pi f = \sqrt{\dfrac{k}{m}}[/tex]
[tex]A = \dfrac{v_m}{2\pi f} = \dfrac{2.26\text{ m/s}}{2\pi\times2.29\text{ Hz}} = 0.157\text{ m}[/tex]
(c) Maximum acceleration is given by
[tex]a = \omega^2A = \dfrac{k}{m}A[/tex]
[tex]a = \dfrac{184\text{ N/m}}{0.885\text{ kg}}\times0.157\text{ m} = 32.6\text{ m/s}^2[/tex]
(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:
[tex]KE_\text{max} = \frac{1}{2}mv_m^2 = \dfrac{1}{2}(0.885\text{ kg})(2.26 \text{ m/s})^2 = 2.26\text{ J}[/tex]
The maximum energy is 2.26 J.
(e) The velocity at any point is
[tex]v = \omega\sqrt{A^2 - x^2}[/tex]
At point x = 4A,
[tex]v = \omega\sqrt{A^2 - (0.4A)^2} = \omega\sqrt{0.84A^2} = \omega A\sqrt{0.84}[/tex]
Putting the values of A and ω,
[tex]\omega = \sqrt{\dfrac{k}{m}} = 14.42\text{ rad/s}[/tex]
[tex]v = (14.42\text{ rad/s})(\text{0.157 m})\sqrt{0.84} = 2.075\text{ m/s}[/tex]
The kinetic energy is
[tex]KE = \dfrac{1}{2}mv^2 = \dfrac{1}{2}(0.885\text{ kg})(2.075\text{ m/s})^2 = 1.91\text{ J}[/tex]
