Q4.9-16E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A mass attached to a spring oscillates with a period of 3 sec. After 2 kg are added, the period becomes 4 sec. Assuming that we can neglect any damping or external forces, determine how much mass was originally attached to the spring. 

Step-by-Step Solution

Answer

Therefore, 2.57 kg mass was originally attached to the spring

Step 1: General form

The Mass–Spring Oscillator:

 

A damped mass-spring oscillator consists of a mass m attached to a spring fixed at one end, as shown in Figure 4.1. Devise a differential equation that governs the motion of this oscillator, taking into account the forces acting on it due to the spring elasticity, damping friction, and possible external influences.

 

Mass–spring oscillator equation:

 $\begin{array}{c}{\mathbf{F}}_{\mathrm{ext}}\mathbf{=}\left[\mathrm{inertia}\right]\mathbf{y}\mathbf{"}\mathbf{+}\left[\mathrm{damping}\right]\mathbf{y}\mathbf{\text{'}}\mathbf{+}\left[\mathrm{stiffness}\right]\mathbf{y}\\ \mathbf{=}\mathbf{my}\mathbf{"}\mathbf{+}\mathbf{by}\mathbf{\text{'}}\mathbf{+}\mathbf{ky}\dots \left(1\right)\end{array}$

   

 The rule for the bounded equation: Just based on stiffness we can decide whether it is bounded or not if stiffness k > 0 then it is bounded and if k < 0 then it is unbounded.

 

Root finding formula:

 

If. ${\mathbf{b}}^2\mathbf{<}\mathbf{4}\mathbf{ac}$Then,$\mathbf{\alpha }\mathbf{=}\mathbf{-}\frac{b}{2a}$ and.$\mathbf{\beta }\mathbf{=}\frac{1}{2a}\sqrt{\mathbf{4}\mathbf{ac}\mathbf{-}{\mathbf{b}}^2}$

Step 2: Evaluate the equation.

Referring to Problem 13: the formula for the quasi period is;

   $\begin{array}{c}\mathbf{P}\mathbf{=}\frac{2\pi }{\beta }\\ \mathbf{=}\frac{4\mathrm{m\pi }}{\sqrt{\mathbf{4}\mathbf{mk}\mathbf{-}{\mathbf{b}}^2}}\dots \left(2\right)\end{array}$

 

In this case, one does not have damping. So, b = 0. Then,

 

$\begin{array}{c}\mathbf{P}\mathbf{=}\frac{4\mathrm{m\pi }}{\sqrt{4\mathrm{mk}-0}}\\ \mathbf{=}\mathbf{2}\mathbf{\pi }\sqrt{\frac{m}{k}}\end{array}$

 

Where m is the mass and k is stiffness.

 

Given that the period is of 3 sec. After 2 kg is added, the period becomes 4 sec.

 

Then${\mathbf{P}}_1\mathbf{=}\mathbf{3}\mathbf{sec}$, and.${\mathbf{P}}_2\mathbf{=}\mathbf{4}\mathbf{sec}$

 

The second mass is for 2 kg heavier than the original mass. So${\mathbf{m}}_2\mathbf{=}{\mathbf{m}}_1\mathbf{+}\mathbf{2}$,.

 

Substitute the values in P one gets,

 

${\mathbf{P}}_1\mathbf{=}\mathbf{2}\mathbf{\pi }\sqrt{\frac{{\mathbf{m}}_1}{k}}\mathbf{,}{\mathbf{P}}_2\mathbf{=}\mathbf{2}\mathbf{\pi }\sqrt{\frac{{\mathbf{m}}_2}{k}}$

 

Divide period 1 by period 2. And substitute the values of the period.

 $\begin{array}{c}\frac{{\mathbf{P}}_1}{{\mathbf{P}}_2}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }\sqrt{\frac{{\mathbf{m}}_1}{k}}}{\mathbf{2}\mathbf{\pi }\sqrt{\frac{{\mathbf{m}}_2}{k}}}\\ \mathbf{=}\frac{\sqrt{{\mathbf{m}}_1}}{\sqrt{{\mathbf{m}}_2}}\\ \frac{3}{4}\mathbf{=}\sqrt{\frac{{\mathbf{m}}_1}{{\mathbf{m}}_1\mathbf{+}\mathbf{2}}}\\ \frac{9}{16}\mathbf{=}\frac{{\mathbf{m}}_1}{{\mathbf{m}}_1\mathbf{+}\mathbf{2}}\\ \mathbf{9}{\mathbf{m}}_1\mathbf{+}\mathbf{18}\mathbf{=}\mathbf{16}{\mathbf{m}}_1\\ \mathbf{7}{\mathbf{m}}_1\mathbf{=}\mathbf{18}\\ {\mathbf{m}}_1\mathbf{=}\frac{18}{7}\approx \mathbf{2}\mathbf{.}\mathbf{57}\end{array}$

So, the original spring was attached to approximately 2.57 kg.