91Ó°ÊÓ

Frequency is defined as the number of oscillations done by an object per second.

The frequency of the mass in SHM is given by

$f=\frac{1}{2\mathrm{Ï€}}\sqrt{\frac{k}{m_f}}$

Here, k is the constant, and$m_f$ is the mass.

The amplitude of motion can be defined as the maximum distance that an object travels before coming back to its original position.

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

The time period is

$\mathrm{T}=\frac{1}{\mathrm{f}}$

Consider the given data as below.

The mass before the collision is ${\mathrm{m}}_{\mathrm{i}}=10\mathrm{kg}$.

Mass after the collision is ${\mathrm{m}}_{\mathrm{f}}=20\mathrm{kg}$.

Speed before the collision is ${\mathrm{v}}_{\mathrm{i}}=2\mathrm{m}/\mathrm{s}$.

Spring force constant is k =170 N/m .

Let consider that the mass of the object before collision is ${\mathrm{m}}_{\mathrm{i}}$, mass of the object after collision is ${\mathrm{m}}_{\mathrm{i}}$, speed of the object before collision is ${\mathrm{v}}_{\mathrm{i}}$, and speed of the object after collision is ${\mathrm{v}}_{\mathrm{f}}$.

Since as the frequency of the mass in SHM is given by

$$\begin{gathered} f=\frac{1}{2\mathrm{Ï€}}\sqrt{\frac{k}{m_f}} \\ =\frac{1}{2\mathrm{Ï€}}\sqrt{\frac{170}{20}} \\ =0.464\mathrm{Hz} \end{gathered}$$

As the relation between frequency and period of SHM is

$$\begin{gathered} T=\frac{1}{f} \\ =\frac{1}{0.464} \\ =2.16s \end{gathered}$$

Now, according to the conservation of momentum, the momentum before collisionrole="math" localid="1668153526208" $p_f$is equal to the momentum after collisioni.e.

${\mathrm{p}}_{\mathrm{i}}={\mathrm{p}}_{\mathrm{f}}$

Or

${\mathrm{m}}_{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}}={\mathrm{m}}_{\mathrm{f}}{\mathrm{v}}_{\mathrm{f}}$

Form the above equation, the speed after collision will be

$\begin{array}{r}{\mathrm{v}}_{\mathrm{f}}=\frac{{\mathrm{m}}_{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}}}{{\mathrm{m}}_{\mathrm{f}}}\\ =\frac{10×2}{20}\\ =1\mathrm{m}â‹\dots {\mathrm{s}}^{−1}\end{array}$

Next, according to the conservation of energy at maximum displacement

x = A

Here, A is amplitude.

The kinetic energy after collision is equal to the elastic potential energy.

$\frac{1}{2}m{v}_f^2=\frac{1}{2}k{A}^2$

By rearranging the above equation, we get

$$\begin{gathered} \mathrm{A}=\sqrt{\frac{{\mathrm{m}}_{\mathrm{f}}{\mathrm{v}}_{\mathrm{f}}}{\mathrm{k}}} \\ =\sqrt{\frac{20×1}{170}} \\ =0.343\mathrm{m} \end{gathered}$$

Because the time taken by the system to return the first time to the position it had immediately after the collision is half of its period i.e

$$\begin{gathered} \mathrm{t}=\frac{\mathrm{T}}{2} \\ =\frac{2.16\mathrm{s}}{2} \\ =1.08\mathrm{s} \end{gathered}$$

Hence, the total time taken to the position after collision is 1.08 s.