91影视

1. Frequency of oscillations,$\mathrm{f}=3.00\mathrm{Hz}$
2. Mass of car,${\mathrm{m}}_{\mathrm{car}}=1450\mathrm{kg}$
3. Mass of each passenger, ${\mathrm{m}}_{\mathrm{passenger}}=73\mathrm{kg}$.

1. The car is mounted on four springs. First, we calculate the mass supported by each spring as we have to calculate the spring constant. We can use the formula for angular frequency in terms of the spring constant and mass supported by each spring to find angular frequency and frequency.
2. To calculate frequency, we first calculate the total mass by considering each passenger鈥檚 mass and the car鈥檚 mass. We can find the mass supported by each spring. We can use the spring constant found in the first part to find the frequency.

Formula:

Frequency relation to the spring constant of a body

$\mathrm{蝇}=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$(i)

The angular frequency of the motion is given as follows:

$\begin{array}{rcl}\mathrm{蝇}& =& 2\mathrm{蟺蹿}\\ & =& 2\mathrm{蟺}脳3\\ & =& 18.85\mathrm{rad}/\sec \end{array}$

Mass supported by each spring is one-fourth of the mass of the car. Hence, the mass load on each spring is given as

$\begin{array}{rcl}\mathrm{m}& =& \frac{1450}{4}\\ & =& 362.5\mathrm{kg}\end{array}$

Now using equation (i), the spring constant can be derived as

$\begin{array}{rcl}18.85& =& \sqrt{\frac{\mathrm{k}}{362.5}}\\ \mathrm{k}& =& 128804.41\mathrm{N}/\mathrm{m}\\ & =& 1.29脳{10}^5\mathrm{N}/\mathrm{m}\end{array}$

Hence, the value of the spring constant is $1.29脳{10}^5\mathrm{N}/\mathrm{m}$.

Now five passengers having an average mass of 73 kg ride inthecar, sothetotal mass is as follows:

$\begin{array}{rcl}\mathrm{M}& =& {\mathrm{m}}_{\mathrm{car}}+5\left({\mathrm{m}}_{\mathrm{passenger}}\right)\\ & =& 1450+5\left(73\right)\\ & =& 1815\mathrm{kg}\end{array}$

So, mass supported by each spring is

$\begin{array}{rcl}\mathrm{m}& =& \frac{1815}{4}\\ & =& 453.75\mathrm{kg}\end{array}$

Now, using equation (i), we get the angular frequency as follows:

$$\begin{gathered} \mathrm{蝇}=\sqrt{\frac{129000}{453.75}} \\ =16.86\mathrm{rad}/\sec \\ 2\mathrm{蟺蹿}=16.86\mathrm{rad}/\sec \left(鈭�\mathrm{angular}\mathrm{frequency}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{frequency},\mathrm{蝇}=2\mathrm{蟺蹿}\right) \\ \mathrm{f}=8.43\mathrm{Hz} \end{gathered}$$

Hence, the frequency of oscillation is $8.43\mathrm{Hz}$.