Question

A 0.23 kg mass at the end of a spring oscillates 2.0 times per second with an amplitude of 0.15 m

Answer 1

Answer:

A) v = 1.885 m/s

B) v = 0.39 m/s

C) E = 0.03 J

D) $x(t) = (0.15m)\cos(2\pi (2.0Hz)t)$

Explanation:

Part A

We will use the conservation of energy to find the speed at equilibrium.

$K_{eq} + U_{eq} = K_A + U_A\\\frac{1}{2}mv^2 + 0 = 0 + \frac{1}{2}kA^2\\v = \sqrt{\frac{k}{m}}A$

where $\omega = \sqrt{k/m}$ and $\omega = 2\pi f$

Therefore,

$v = 2\pi f A = 2(3.14)(2)(0.15) = 1.885~m/s$

Part B

The conservation of energy will be used again.

$K_1 + U_1 = K_2 + U_2\\\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2\\mv^2 + kx^2 = kA^2\\(0.23)v^2 + k(0.10)^2 = k(0.15)^2\\v^2 = \frac{k(0.15)^2-(0.10)^2}{0.23}\\v = \sqrt{0.054k}$

where $k = \omega^2 m = (2\pi f)^2 m = 2(3.14)(2)(0.23) = 2.89$

Therefore, v = 0.39 m/s.

Part C

Total energy of the system is equal to the potential energy at amplitude.

$E = \frac{1}{2}kA^2 = \frac{1}{2}(2.89)(0.15)^2 = 0.03~J$

Part D

The general equation of motion in simple harmonic motion is

$x(t) = A\cos(\omega t + \phi)\\x(t) = (0.15m)\cos(2\pi (2.0Hz)t + \phi)$

where $\phi$ 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,

$x(t) = (0.15m)\cos(2\pi (2.0Hz)t)$