91Ó°ÊÓ

• Frequency (f)=2.00 Hz.
• Length of the spring x=12.0 cm or 0.12m.
• Reading of scale balance ranges from m =0 or 15 kg.

A particle with mass m that moves under the influence of Hooke’s law restoring force, $\mathit{F}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathit{k}\mathit{x}$,exhibits simple harmonic motion. Here, $\mathit{F}$is restoring force, $\mathit{k}$is force constant and $\mathit{x}$ is the displacement from the mean position.

By using the formula for force constant k and period T, we can find the spring constant and weight of the package.

Formula:

The spring constant of the oscillations, k=F / x (i)

The period of the oscillations, T=1/f or T =2$\mathrm{Ï„}\mathrm{Ï„}\sqrt{\left(\frac{M}{k}\right)}$ (ii)

The weight of a body, W=mg (iii)

Using equation (i), the spring constant of an oscillation can be given as:

$$\begin{gathered} k=\frac{mg}{x}\left(âˆ\mu \mathrm{Force}\mathrm{acting}\mathrm{on}\mathrm{the}\mathrm{body},F=mg\right) \\ =\frac{15\mathrm{kg}×9.8\mathrm{m}/{\mathrm{s}}^2}{0.12\mathrm{m}} \\ =1225\mathrm{N}/\mathrm{m} \end{gathered}$$

Hence, the value of spring constant is 1225 N/m

The time period of a body using equation (ii) can be given as:

$$\begin{gathered} T=\frac{1}{2} \\ =0.5s \end{gathered}$$

Now, the mass of the package can be given using equation (ii) as:

By squaring and rearranging for M

$$\begin{gathered} M=\frac{T^{2}k}{4\mathrm{τ}{\mathrm{τ}}^2} \\ =\frac{{\left(0.5\mathrm{s}\right)}^2×1225\mathrm{N}/\mathrm{m}}{4{\left(3.14\right)}^2} \\ =7.757\mathrm{kg} \end{gathered}$$

$\mathrm{So},\mathrm{the}\mathrm{weight}\mathrm{of}\mathrm{the}\mathrm{package}\mathrm{using}\mathrm{equation}\left(\mathrm{iii}\right)\mathrm{can}\mathrm{be}\mathrm{given}\mathrm{as}:$

$$\begin{gathered} \mathrm{W}=7.757\mathrm{kg}×9.8\mathrm{m}/{\mathrm{s}}^2 \\ =76.0\mathrm{N} \end{gathered}$$

$\mathrm{Hence},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{weight}\mathrm{is}76.0\mathrm{N}.$