Question

A hanging spring stretches by $35.0\text{cm}$ when an object of mass $450\text{g}$ is hung on it at rest. In this situation, we define its position as $x=0$. The object is pulled down an additional $18.0\text{cm}$ and released from rest to oscillate. Find the position of the vibrating object from $x=0$ after $84.4\text{s}$ $($Neglect friction due to air$)$

• A. $10.2\text{cm}$
• B. $15.8\text{cm}$
• C. $7\text{cm}$
• D. $5.8\text{cm}$

Answer 1

The correct option is B $15.8\text{cm}$

Given,
Mass of the object, $m=450\text{g}=0.45\text{kg}$
Change in length when mass $m$ is hanged, $x=35\text{cm}=0.35\text{m}$

The spring constant of this spring is given by
$k=\frac{F}{x}$
where, $F$ is the tensile force acting on the spring due to weight of the obejct.

$\Rightarrow k=\frac{mg}{x}$
Substituting the values ,

$\Rightarrow k=\frac{\left(0.450\text{kg}\right)\left(9.80{\text{m/s}}^2\right)}{0.350\text{m}}$

$\Rightarrow k=12.6\text{N/m}$

The block mass system will perform simple harmonic motion when it is released after stretching $18\text{cm}$ from equilibrium position.

We take the $\text{x-axis}$ pointing downward, so the phase angle $\varphi =0$.
Thus, the position of block after $84.4\text{s}$ is given by
$x=A\cos \omega t$

Substituting the values,

$\Rightarrow x=\left(18.0\text{cm}\right)\cos \left[\sqrt{\frac{12.6\text{N/m}}{0.450\text{kg}}}\right(84.4\text{s}\left)\right]$

$\Rightarrow x=15.8\text{cm}$

Hence, option (b) is correct answer.