Question

On your first trip to Planet $X$you happen to take along a $200g$mass, a $40cm$-long spring, a meter stick, and a stopwatch. You’re curious about the free-fall acceleration on Planet $X$, where ordinary tasks seem easier than on earth, but you can’t find this information in your Visitor’s Guide. One night you suspend the spring from the ceiling in your room and hang the mass from it. You find that the mass stretches the spring by $31.2cm$. You then pull the mass down $10.0cm$and release it. With the stopwatch you find that $10$oscillations take $14.5s$. Based on this information, what is $g$?

Short answer

The value of $g$on planet is $X$localid="1650092781874" $5.85\mathrm{m}/{\mathrm{s}}^2$

Step-by-step solution

Concepts and Principles

The period of oscillation in simple harmonic motion is given by $T=2\pi \sqrt{\frac{m}{k}}$

$T$is independent on amplitude but only on the mass $m$and the force constant$k$

Given data

The mass of the object is $m=\left(200\mathrm{g}\right)\left(\frac{1\mathrm{kg}}{1000\mathrm{kg}}\right)=0.200\mathrm{kg}$

The distance stretched in the spring by mass when the spring is hung from ceiling$\mathrm{\Delta x}=(31.2\mathrm{cm})\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.312\mathrm{m}$

The time of $10$oscillation of the mass is$\mathrm{\Delta }t=14.5\mathrm{s}$

Required data

The objective is to determine the value of $g$on planet$X$

Solution

Figure 1 displays the free body diagram of the forces acting on the mass when the spring is hung from the ceiling; $\overrightarrow{\mathbf{k}}\mathrm{\Delta }x$is the spring force and $m\overrightarrow{\mathit{g}}$is the gravitational force exerted by earth

Solve

As the mass in at rest, so the net force on it must be zero

$\sum F_y=k\mathrm{\Delta }x-mg=0$

$k\mathrm{\Delta }x=mg$

To find $k$

$k=\frac{mg}{\mathrm{\Delta }x}$

substituting values

$\begin{array}{r}g=\left(\frac{0.312\mathrm{m}}{0.200\mathrm{kg}}\right)k\end{array}$

$\begin{array}{r}=(1.56\mathrm{m}/\mathrm{kg})k\end{array}$

Solve

The period of oscillation of the mass can be obtained from equation $\left(1\right)$

$T=2\pi \sqrt{\frac{m}{k}}$

where $T=\mathrm{\Delta t}/10$on the behalf of stop watch measures $10$oscillations

$\frac{\mathrm{\Delta }t}{10}=2\pi \sqrt{\frac{m}{k}}$

Rearrange and solve for $k$

$\begin{array}{r}{\left(\frac{\mathrm{\Delta }t}{10}\right)}^2=4{\pi }^2\frac{m}{k}\\ k=4{\pi }^{2}m{\left(\frac{10}{\mathrm{\Delta }t}\right)}^2\end{array}$

Substitute $k$value in equation $\left(2\right)$

$\frac{1.56\mathrm{m}/\mathrm{kg}}{g}=4{\pi }^{2}m{\left(\frac{10}{\mathrm{\Delta }t}\right)}^2$

solve  

Solve for $g$

$g=4{\pi }^{2}m(1.56\mathrm{m}/\mathrm{kg}){\left(\frac{10}{\mathrm{\Delta }t}\right)}^2$

substituting values in it to find out$g$value

$\begin{array}{r}g=4{\pi }^2\left(0.200\mathrm{kg}\right)(1.56\mathrm{m}/\mathrm{kg})\left(\frac{10}{14.5\mathrm{s}}\right)\end{array}$

$\begin{array}{r}=5.85\mathrm{m}/{\mathrm{s}}^2\end{array}$