Respuesta :
Answer:
A) v = 1.885 m/s
B) v = 0.39 m/s
C) E = 0.03 J
D) [tex]x(t) = (0.15m)\cos(2\pi (2.0Hz)t)[/tex]
Explanation:
Part A
We will use the conservation of energy to find the speed at equilibrium.
[tex]K_{eq} + U_{eq} = K_A + U_A\\
\frac{1}{2}mv^2 + 0 = 0 + \frac{1}{2}kA^2\\
v = \sqrt{\frac{k}{m}}A[/tex]
where [tex]\omega = \sqrt{k/m}[/tex] and [tex]\omega = 2\pi f[/tex]
Therefore,
[tex] v = 2\pi f A = 2(3.14)(2)(0.15) = 1.885~m/s[/tex]
Part B
The conservation of energy will be used again.
[tex]K_1 + U_1 = K_2 + U_2\\
\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2\\
mv^2 + kx^2 = kA^2\\
(0.23)v^2 + k(0.10)^2 = k(0.15)^2\\
v^2 = \frac{k(0.15)^2-(0.10)^2}{0.23}\\
v = \sqrt{0.054k}[/tex]
where [tex]k = \omega^2 m = (2\pi f)^2 m = 2(3.14)(2)(0.23) = 2.89[/tex]
Therefore, v = 0.39 m/s.
Part C
Total energy of the system is equal to the potential energy at amplitude.
[tex]E = \frac{1}{2}kA^2 = \frac{1}{2}(2.89)(0.15)^2 = 0.03~J[/tex]
Part D
The general equation of motion in simple harmonic motion is
[tex]x(t) = A\cos(\omega t + \phi)\\
x(t) = (0.15m)\cos(2\pi (2.0Hz)t + \phi)[/tex]
where [tex]\phi[/tex] is the phase angle to be determined by the initial conditions. In this case, the initial condition is that at t = 0, x is maximum. Therefore,
[tex]x(t) = (0.15m)\cos(2\pi (2.0Hz)t)[/tex]
