Question

A mass of $\mathbf{2}\mathbf{.}\mathbf{2}\mathbf{}\mathit{k}\mathit{g}$is connected to a horizontal spring whose stiffness is role="math" localid="1657769591862" $\mathbf{8}\mathbf{}\mathit{N}\mathbf{/}\mathit{m}$. When the spring is relaxed,$\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}$ The spring is stretched so that the initial value of $\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{18}\mathbf{}\mathit{m}$. The mass is released from rest at time role="math" localid="1657769734196" $\mathit{t}\mathbf{=}\mathbf{}\mathbf{0}$. Remember that when the argument of a trigonometric function is in radians, on a calculator you have to switch the calculator to radians or convert the radians to degrees. Predict the position x when$\mathit{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{15}\mathbf{}\mathit{s}$.

Short answer

The displacement of spring is$x=-0.1049m$when at $t=1.15s$.

Step-by-step solution

Identification of given data

Given data can be listed below,

• Mass,$m=2.2kg$

• Spring stiffness, $k=8N/m$

• Displacement of spring, $x=0.18m$

Evaluating the angular frequency of the spring-mass system

Expression for angular frequency of the spring-mass system is given by

$$\begin{gathered} \omega =\sqrt{\frac{k}{m}} \\ =\sqrt{\frac{8}{2.2}} \\ =1.907rad/s \end{gathered}$$

Displacement of spring when t=1.15 m

Expression for analytical solution for spring-mass system is given by

$\omega t=A\cos ()$

The spring is stretched initial value of $x=0.18m$When time $t=0$

Substituting $x=1.18m$and $t=0$in the above equation we get

$$\begin{gathered} 0.18=A\cos (1.09\times 0) \\ 0.18=A \\ A=0.18m \\ Whent=1.15s \\ x=0.18\cos \left(1.907rad\times \frac{{180}^0}{\pi }\times 1.15s\right) \\ =-0.1049m \end{gathered}$$

Thus, the displacement of spring $x=-0.1049m$when $t=1.15s$