Question

Compact car has a mass of 1260 kgkg . Assume that the car has one spring on each wheel, that the springs are identical, and that the mass is equally distributed over the four springs.

A. What is the spring constant of each spring if the empty car bounces up and down 1.6 times each second?Express your answer using two significant figures.in N/m.B. What will be the car's oscillation frequency while carrying four 70 kg passengers?Express in two sig figs in Hz.

Answer 1

The spring constant of each spring in the compact car is approximately 6.53 x 10^4 N/m. The car's oscillation frequency while carrying four 70 kg passengers is approximately 1.37 Hz.

Step-by-step explanation:

To calculate the spring constant of each spring, we can use Hooke's Law, which states that the force applied by a spring is directly proportional to the displacement from its equilibrium position. In this case, the displacement is given as 1.20 cm (or -1.20 × 10^-2 m) and the force is the weight of the car, which is equal to 784 N. We can use the equation F = -kx, where F is the force, k is the spring constant, and x is the displacement. Rearranging the equation to solve for k, we have k = -F/x. Plugging in the values, k = -784 N / (-1.20 × 10^-2 m) = 6.53 x 10^4 N/m. Therefore, the spring constant of each spring is approximately 6.53 x 10^4 N/m.

To calculate the oscillation frequency of the car carrying four 70 kg passengers, we need to consider the total mass of the car and passengers. The mass of the car is given as 1260 kg and each passenger weighs 70 kg, so the total mass is 1260 kg + (4 * 70 kg) = 1540 kg. The oscillation frequency can be calculated using the formula f = 1 / (2π) * sqrt(k / m), where f is the frequency, k is the spring constant, and m is the mass. Plugging in the values, f = 1 / (2π) * sqrt((6.53 x 10^4 N/m) / 1540 kg) ≈ 1.37 Hz.

Answer 2

given,

mass of the car, m = 1260 Kg

mass on each spring = 1260/4 = 315 Kg

a) frequency, f = 1.6 Hz

spring constant, k = ?

`f = (1)/(2\pi) \sqrt{(k)/(m)}`

squaring both side

`k = 4 \pi^2 f^2 m`

`k = 4 \pi^2* 1.6^2* 315`

k = 3.18 x 10⁴ N/m

b) frequency of the spring if 4 mass of 70 Kg sit in the car

mass on each spring,
`m = (1260 + 4* 70)/(4)`

m = 385 Kg

now,

`f = (1)/(2\pi) \sqrt{(k)/(m)}`

`f = (1)/(2\pi) \sqrt{(3.18* 10^(4))/(385)}`

f = 1.45 Hz

hence, frequency of the spring is equal to 1.45 Hz