Question

Suppose a spring-mass system has 18Nm 1 −
k = and m = 0.71kg. The system is
oscillating with an amplitude of 54 mm. (i) Determine the angular frequency of
oscillation. (ii) Obtain an expression for the velocity v of the block as a function of
displacement, x and calculate v at x = 34 mm. (iii) Obtain an expression for the mass’s
distance | x | from the equilibrium position as a function of the velocity v and calculate | x |
when v = 0.18 ms−1. (iv) Calculate the energy of the spring-mass system.

Answer 1

QUESTION:

Suppose a spring-mass system has $18\mathrm{Nm}$ $k =$ and $m = 0.71\mathrm{kg}$. The system is oscillating with an amplitude of $54\mathrm{mm}$. (i) Determine the angular frequency of oscillation.

(ii) Obtain an expression for the velocity $\mathbf{v}$ of the block as a function of displacement, $x$ and calculate $\mathbf{v}$ at $x = 34$ mm. (iii) Obtain an expression for the mass's distance $|x|$ from the equilibrium position as a function of the velocity $\mathbf{v}$ and calculate $|x|$ when $\mathbf{v} = 0.18$ ms⁻¹. (iv) Calculate the energy of the spring-mass system.

SOLUTION:

The angular frequency of oscillation $\omega = \sqrt{\frac{k}{m}} = 5.05\frac{\mathrm{rad}}{\mathrm{s}}$

The velocity of the block is $\upsilon(t) = \frac{dx}{dt} = \frac{d}{dt} \left( x_0 \sin(\omega t + \phi_0) \right) = \omega x_0 \cos(\omega t + \phi_0)$

Since $\cos^2\alpha + \sin^2\alpha = 1$, $x_0 \cos(\omega t + \phi_0) = x_0 \sqrt{1 - \sin^2(\omega t + \phi_0)} = \sqrt{x_0^2 - x(t)^2}$, therefore $\upsilon(t) = \omega \sqrt{x_0^2 - x(t)^2} \cdot x_0 = 54 \, \mathrm{mm}$ — amplitude of oscillation

When $x = 34 \, \mathrm{mm}$, $\upsilon = \omega \sqrt{x_0^2 - x^2} = 0.2119 \, \mathrm{m/s}$

$$x(t) = x_0 \sin(\omega t + \phi_0) = x_0 \sqrt{1 - \cos^2(\omega t + \phi_0)} = x_0 \sqrt{1 - \left(\frac{\upsilon(t)}{\omega \cdot x_0}\right)^2}$$

When $\upsilon = 0.18 \, \mathrm{m/s}$

$$x(t) = 40.56 \, \mathrm{mm}$$

The energy of the system is $E = \frac{k \cdot x_0^2}{2} = 0.0262 \, \mathrm{J}$