Question

A certain mass stretches one spring $\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{3}$}\right.\mathbf{\text{foot}}$and another spring $\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.\mathbf{\text{foot}}$. The two springs are then attached in parallel to a common rigid support in the manner shown in Figure 5.1.5. The first mass is set aside, and a mass weighing $\mathbf{\text{8pounds}}$is attached to the double-spring arrangement, and the system is set in motion. If the period of motion is $\raisebox{1ex}{$\mathbf{\pi }$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{15}}$}\right.\mathbf{\text{second}}$, determine how much the first mass weighs.

Short answer

The first mass weight is $W_0=45\text{lb}$.

Step-by-step solution

Hooke’s law:

Hooke's law is a law of physics that states that the force needed to extend or compress a spring by some distance scales linearly with respect to that distance that is,

$\mathbf{}\mathit{F}\mathbf{=}\mathit{k}\mathit{x}$

here, $\mathit{k}$is a constant factor characteristic of the spring and $\mathit{x}$is small compared to the total possible deformation of the spring.

Find spring constant:

Let the weight of the first mass be $W_0$.

Given $x_1=\frac{1}{3}$and $x_2=\frac{1}{2}$

here,$x_1$and$x_2$are the displacement of the two strings.

Using Hooke's Law find the spring constants of the two strings.

$\begin{array}{rcl}{k}_1& =& \frac{W_0}{x_1}\\ & =& \frac{W_0}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\\ & =& 3W_0\end{array}$

And

$\begin{array}{rcl}{k}_2& =& \frac{W_0}{x_2}\\ & =& \frac{W_0}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ & =& 2W_0\end{array}$

Effective spring constant:

The effective spring constant is,

$\begin{array}{rcl}K& =& k_1+k_2\\ & =& 3W_0+2W_0\\ & =& 5W_0\end{array}$

The given data:

If $x\left(t\right)$is the displacement from the equilibrium position, then Newton's second law is given by,

$\frac{d^{2}x}{d{t}^2}+\frac{k}{m}x=0$ ..... (1)

Given a mass weighing $8\text{pounds}$is attached to the double spring that is,

localid="1668405083046" $mg=8\text{lb}$

here,localid="1668405095777" $g$is gravity measured as localid="1668405128202" $32\raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.$.

Then;

localid="1668405155726" $\begin{array}{rcl}m& =& \frac{8}{32}\frac{\text{lb}}{\raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.}\\ & =& \frac{1}{4}\text{slug}\end{array}$

Therefore, localid="1668405168212" $m=\frac{1}{4}\text{slug}$slug.

Sincelocalid="1668405208818" $\text{1slug}=1\frac{\text{lb}\cdot {\text{s}}^{\text{2}}}{\text{ft}}$

Find angular velocity:

Substitute $K=5W_0$and $m=\frac{1}{4}\text{slug}$into equation (1) then

$\begin{array}{rcl}\frac{d^{2}x}{d{t}^2}+\frac{k}{m}x& =& 0\\ \frac{d^{2}x}{d{t}^2}+\frac{5W_0}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}x& =& 0\\ \frac{d^{2}x}{d{t}^2}+20W_{0}x& =& 0\end{array}$

Let${\omega }^2=20W_0$then$\frac{d^{2}x}{d{t}^2}+20W_{0}x=0$can be written as

$$\begin{gathered} x^{\text{'}\text{'}}+{\omega }^{2}x=0 \\ x\left(t\right)=c_1\cos \omega t+c_2\sin \omega t \end{gathered}$$

Since${\omega }^2=20W_0$then $\omega =\sqrt{20W_0}$.

Find time period:

The period of motion is,

$\begin{array}{rcl}T& =& \frac{2\pi }{\omega }\\ & =& \frac{2\pi }{\sqrt{20W_0}}\end{array}$

Given period of motion is $\frac{\pi }{15}\text{second}$then,

$\begin{array}{rcl}T& =& \frac{2\pi }{\sqrt{20W_0}}\\ \frac{\pi }{15}& =& \frac{2\pi }{\sqrt{20W_0}}\\ \sqrt{20W_0}& =& 30\end{array}$

Squaring on both sides and simplifying you get.

$20W_0=900$

$\begin{array}{rcl}{W}_0& =& \frac{900}{20}\\ & =& 45\text{lb}\end{array}$

Hence, the first mass weight is $45\text{lb}$.