91Ó°ÊÓ

The period can be defined as the time taken required to complete one cycle or oscillation.

Calculate the masses.

Consider the given data as below.

Mass of the bucket, ${\mathrm{m}}_{\mathrm{b}}=2\mathrm{kg}$

Mass of water,$m_w=10kg$

The initial mass of the system is

$$\begin{gathered} \mathrm{M}={\mathrm{m}}_{\mathrm{b}}+{\mathrm{m}}_{\mathrm{w}} \\ =10+2 \\ =12\mathrm{kg} \end{gathered}$$

Now, when the bucket is half full, then the mass of water will be half and mass of the system will be

$\begin{array}{r}{M}^{\mathrm{′}}=m_b+\frac{m_w}{2}\\ =2+5\\ =7\mathrm{kg}\end{array}$

The period of oscillations in S.H.M. is given by

$\begin{array}{r}\mathrm{T}=2\mathrm{Ï€}\sqrt{\frac{\mathrm{M}}{\mathrm{k}}}\\ \mathrm{T}=2\mathrm{Ï€}\sqrt{\frac{7}{450}}\\ =0.784\mathrm{s}\end{array}$

The time period is given by

$T=2\mathrm{Ï€}\sqrt{\frac{{\mathrm{m}}_{\mathrm{b}}+{\displaystyle \frac{{\mathrm{m}}_{\mathrm{w}}}{2}}}{\mathrm{k}}}$

Differentiate the above equation with respect to time to get the rate of change of time period

$$\begin{gathered} \frac{\mathrm{dT}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left[2\mathrm{π}\sqrt{\frac{{\mathrm{m}}_{\mathrm{b}}+\frac{{\mathrm{m}}_{\mathrm{w}}}{2}}{\mathrm{k}}}\right] \\ \frac{\mathrm{dT}}{\mathrm{dt}}=2\mathrm{π}\sqrt{\frac{1}{\mathrm{k}}}\frac{\mathrm{d}}{\mathrm{dt}}\sqrt{{\mathrm{m}}_{\mathrm{b}}+\frac{{\mathrm{m}}_{\mathrm{w}}}{2}} \\ =2\mathrm{π}\sqrt{\frac{1}{\mathrm{k}}×\frac{1}{2}{\left({\mathrm{m}}_{\mathrm{b}}+\frac{{\mathrm{m}}_{\mathrm{w}}}{2}\right)}^{−\frac{1}{2}}\frac{{\mathrm{dmw}}_{\mathrm{w}}}{\mathrm{dt}}} \\ =2\mathrm{π}\sqrt{\frac{1}{450}}×\frac{1}{2}{\left(2+\frac{10}{2}\right)}^{−\frac{1}{2}}×−2×{10}^{−3} \\ =−1.12×{10}^{−4} \end{gathered}$$

Since the period is directly proportional to the square root of mass of the system and the mass of the system is decreasing. Hence, the period is getting shorter.

The period will be shortest when ${\mathrm{m}}_{\mathrm{w}}=0$. Hence, the shortest time period will be

$$\begin{gathered} T=2\mathrm{Ï€}\sqrt{\frac{m_b+0}{k}} \\ =2\mathrm{Ï€}\sqrt{\frac{m_b}{k}} \\ =2\mathrm{Ï€}\sqrt{\frac{2}{450}} \\ =0.42\mathrm{s} \end{gathered}$$

Hence, the shortest period is T =0.42 s.