91Ó°ÊÓ

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

The period of the empty car will be

${\mathbf{T}}_{\mathbf{E}}\mathbf{=}\mathbf{2}\mathbf{Ï„Ï„}\sqrt{\frac{\mathbf{m}}{\mathbf{k}}}$

Here, ${\mathbf{T}}_{\mathbf{E}}$ is the time period of the empty car, m is the mass of the car, and k is the spring constant.

Consider the given data as below.

The combined mass of the passengers is M =250 kg .

Period of vibration of the loaded car is ${\mathrm{T}}_{\mathrm{L}}=1.92\mathrm{s}$.

Let us consider here that the mass of the car is , and the period of vibration of the empty car is ${\mathrm{T}}_{\mathrm{E}}$.

The period of the object of mass and a spring having a spring constant k is given by

$\mathrm{T}=2\mathrm{ττ}\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$ ….. (1)

Since the mass of the combined system i.e. car and passengers is (m+M) , so the period of the loaded system will be

${\mathrm{T}}_{\mathrm{L}}=2\mathrm{ττ}\sqrt{\frac{\mathrm{m}+\mathrm{M}}{\mathrm{k}}}$ ….. (2)

And the period of the empty car will be

${\mathrm{T}}_{\mathrm{E}}=2\mathrm{ττ}\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$ ….. (3)

Now divide the equation (3) by (2) as below.

$$\begin{gathered} \frac{T_E}{T_L}=\frac{2\mathrm{π}\sqrt{\frac{m}{k}}}{2\mathrm{π}\sqrt{\frac{m+M}{k}}} \\ =\sqrt{\frac{m}{m+M}} \\ {T}_E=T_L\sqrt{\frac{m}{m+M}} \end{gathered}$$ ….. (4)

First calculate the spring force constant.

Here given that the displacement in the spring by the mass of 250 kg, since this mass is balanced by the spring force i.e.

Mg =kx ….. (5)

Here, g is the acceleration due to gravity and x is the displacement.

Consider the given data as below.

Acceleration due to gravity,$\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2$

Mass of the passengers, $\mathrm{M}=250\mathrm{kg}$

The displacement, $\mathrm{x}=4.0\mathrm{cm}=4.0×{10}^{-2}\mathrm{m}$

By rearranging the equation (5), you obtain

$$\begin{gathered} \mathrm{k}=\frac{\mathrm{Mg}}{\mathrm{x}} \\ =\frac{250×9.8}{4×{10}^{−2}} \\ =6.125×{10}^3\mathrm{N}â‹\dots {\mathrm{m}}^{−1} \end{gathered}$$

Hence by rearranging the equation (2) as below, you will get

$\mathrm{m}=\mathrm{k}\left(\frac{{\mathrm{T}}_{\mathrm{L}}^2}{4{\mathrm{Ï€}}^2}\right)-\mathrm{M}$

Substitute all known values in the above equation.

$$\begin{gathered} \mathrm{m}=6.125×{10}^3×\frac{{\left(1.92\right)}^2}{4{\left(3.14\right)}^2}-250 \\ =5.47×{10}^3\mathrm{kg} \end{gathered}$$

Hence the period of the empty car define by substituting known values into equation (4).

$$\begin{gathered} {\mathrm{T}}_{\mathrm{E}}=1.92×\sqrt{\frac{5.47×{10}^3}{5.47×{10}^3+250}} \\ =1.88\mathrm{s} \end{gathered}$$

Hence, the time period of the empty car is 1.88 s.