A compact car has a mass of 1150 kg. 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 2.2 times each second?
(b) What will be the car's oscillation frequency while carrying four 80 kg passengers?
2PI*freq= sqrt (k/m)
a. the mass on each spring is 1150/4,
(2PI*2.2)^2*1150/4= k
b. freq= 1/2PI * sqrt (kabove/((
1150/4 + 80)
(a) Well, it seems like the car is really good at bouncing! To find the spring constant, we'll need to use Hooke's law, which says that the force exerted by a spring is proportional to the displacement from its equilibrium position. In this case, the equilibrium position is when the car is not bouncing.
The formula for the spring constant (k) is given by the equation:
k = (4π^2 * m * f^2) / (4 * h^2)
Where:
- k is the spring constant,
- m is the mass of the car (1150 kg),
- f is the frequency of the car bouncing up and down (2.2 Hz), and
- h is the amplitude of the car's oscillations (which we'll assume is the same as the maximum displacement).
Plugging in these values into the equation, we get:
k = (4π^2 * 1150 kg * (2.2 Hz)^2) / (4 * (2h)^2)
Simplifying this, we find:
k = (π^2 * 1150 kg * (2.2 Hz)^2) / (h^2)
(b) Now, let's add some passengers to our bouncing car. Assuming each passenger weighs 80 kg, we now have an additional mass of 4 * 80 kg = 320 kg.
When the car is carrying the passengers, the mass to be distributed over the four springs increases to 1150 kg + 320 kg = 1470 kg.
To find the new oscillation frequency, we can use the same formula as before, but with the updated mass:
f' = sqrt(k / m')
Where:
- f' is the new oscillation frequency,
- k is the spring constant (which we found in part a),
- m' is the new mass of the car (1470 kg).
Plugging in these values, we get:
f' = sqrt(k / m') = sqrt((π^2 * 1150 kg * (2.2 Hz)^2) / (h^2) / 1470 kg)
Now you can solve for f' and enjoy your bouncing ride! Just remember to buckle up and don't let your hair get too messy!
(a) To find the spring constant of each spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
The formula for the oscillation frequency, also known as the angular frequency (ω), is given by:
ω = 2πf
where f is the frequency in Hz.
Given that the car bounces up and down 2.2 times each second, the frequency (f) is 2.2 Hz.
We also know that the mass of the car (m) is 1150 kg and it is equally distributed over the four springs.
To calculate the spring constant (k), we can use the formula:
k = (mω²) / 4
Substituting the values, we get:
k = (1150 kg * (2π * 2.2 Hz)²) / 4
Now, let's calculate the spring constant.
k = (1150 kg * 4π² * 2.2² Hz²) / 4
k ≈ 2448 N/m
Therefore, the spring constant of each spring is approximately 2448 N/m.
(b) When the car is carrying four 80 kg passengers, the total mass (m) will be:
m = 1150 kg + (4 * 80 kg)
m = 1150 kg + 320 kg
m = 1470 kg
We can use the same formula to calculate the new spring constant (k).
k = (1470 kg * (2π * f)²) / 4
Since we are looking for the new oscillation frequency, we can rearrange the formula as:
f = √((4k) / (m * 4π²))
Substituting the values:
f = √((4 * 2448 N/m) / (1470 kg * 4π²))
f ≈ √(9792 N/m) / (1470 kg * 39.48)
f ≈ √(2.487 (1/s²) / (0.169 (kg/s²))
f ≈ √14.7 (1/s²) (kg/s²)⁻¹
f ≈ 3.83 Hz
Therefore, the car's oscillation frequency while carrying four 80 kg passengers will be approximately 3.83 Hz.
To find the spring constant of each spring, we need to use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position.
(a) To find the spring constant, we can start by calculating the effective mass that each spring supports. Since the mass is equally distributed over the four springs, each spring supports 1/4th of the total mass.
Effective mass supported by each spring = (mass of the car) / (number of springs)
= 1150 kg / 4
= 287.5 kg
Next, we need to find the angular frequency (ω) of the car's oscillation. The angular frequency is related to the frequency (f), which is given as 2.2 Hz, by the formula:
ω = 2πf
ω = 2π × 2.2 rad/s
Now, let's use Hooke's Law to find the spring constant (k). For a given mass and angular frequency:
ω = √(k / m)
Rearranging the formula to solve for k:
k = m × ω^2
Substituting the values into the formula:
k = 287.5 kg × (2π × 2.2 rad/s)^2
Calculating this will give us the spring constant of each spring.
(b) When the car carries four 80 kg passengers, the total mass supported by each spring will increase.
Total mass supported by each spring = mass of the car + (number of passengers × mass of each passenger)
= 1150 kg + (4 × 80 kg)
= 1150 kg + 320 kg
= 1470 kg
Now, we can repeat the calculation to find the new oscillation frequency of the car. Using the mass of 1470 kg instead of 287.5 kg in the formula:
k = 1470 kg × (2π × 2.2 rad/s)^2
This will give us the new oscillation frequency of the car while carrying four passengers.