Answer to Question #145088 in Physics for Joshua

Question

A 0.350 kg sailplane attached to the end of an ideal spring with force constant 375 N/m, undergoes a simple harmonic motion with an amplitude of 0.030 m. A) Find the speed of the sailplane when it is at x = -0.013 m B) What is the maximum speed of sailplane?

Answer 1

The position of the sailplane involven into a simple harmonic motion is given by:

$$x(t) = A\sin\left(\sqrt{\dfrac{k}{m}} t\right)$$

where $A = 0.03m$ is the amplitude, $k = 375N/m$ is spring constant, $m = 0.35kg$ is the mass of the sailplane and $t$ is a time.

The sailplane has position $x_1 = -0.013m$ at the time:

$$t_1 = \sqrt{\dfrac{m}{k}}\arcsin\left(\dfrac{x_1}{A}\right) \\ t_1 = \sqrt{\dfrac{0.35}{375}}\arcsin\left(\dfrac{-0.013}{0.03}\right) \approx 0.18s$$

The velocity is given by:

$$v(t_1) = x'(t_1) = A\sqrt{\dfrac{k}{m}}\cos\left(\sqrt{\dfrac{k}{m}} t_1\right)\\ v(t_1) = 0.03\sqrt{\dfrac{375}{0.35}}\cos\left(\sqrt{\dfrac{375}{0.35}}\cdot 0.18\right) \approx 0.88m/s$$

The maximum speed is reached when $t = 0$. Then

$$v_{max} = A\sqrt{\dfrac{k}{m}} = 0.03\sqrt{\dfrac{375}{0.35}} \approx 0.98m/s$$

Answer. A) 0.88m/s, B) 0.98m/s.