An automobile can be considered to be mounted on four identical springs as far as vertical oscillations are concerned. The springs of a certain car are adjusted so that the oscillations have a frequency of 3 Hz.
(a) What is the spring constant of each spring if the mass of the car is 1400 kg and the weight is evenly distributed over the springs?
N/m
(b) What will be the vibration frequency if five passengers, averaging 71.0 kg each, ride in the car with an even distribution of mass?
Hz
I will be happy to critique your thinking.
b) k =mw^2, k = 1400 * 18.84^2, k = 496924/4 = 124231 N/M
Christiaan your answer for the spring constant is wrong
not clear enough you have to demostrate all calculation
for a)you must first calculate the angular frequency,w, you are only given the frequency,f. to do that you use the equation:
w=2*pi*f
=2*pi*3Hz
next you take that and put it into the equation with k being the spring constant,w being the angular velocity, and m being the mass:
w= sqrt( k / m )
6*pi= sqrt( k / 1400)
...but since there are 4 springs you need to modify the equation to be:
6*pi= sqrt( 4k / 1400)
so the answer should be 124357.0155
To find the spring constant of each spring, you need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement it undergoes.
(a) To find the spring constant, we first need to determine the effective mass of the car. Since the weight is evenly distributed over the four springs, each spring supports a quarter of the total weight. The weight of the car can be calculated using the equation:
Weight = mass × gravitational acceleration
Weight = 1400 kg × 9.8 m/s² (acceleration due to gravity)
Weight = 13720 N
Since there are four springs, each spring will support a weight of 1/4 of the total weight:
Weight per spring = 13720 N / 4
Weight per spring = 3430 N
Now let's consider a single spring-mass system. The frequency of oscillation (f) is related to the spring constant (k) and the mass (m) by the equation:
f = 1 / (2π) * √(k / m)
Rearranging the equation, we can solve for the spring constant:
k = (2πf)² * m
Plugging in the values:
k = (2π * 3 Hz)² * 3430 N / kg
k ≈ 64186 N/m
Therefore, the spring constant of each spring is approximately 64186 N/m.
(b) In this case, we need to take into account the additional mass of the five passengers. Since the mass is evenly distributed, it means each spring will support the weight of the car plus the weight of the passengers.
Weight per spring = (Weight of the car + Total weight of passengers) / 4
Weight per spring = (13720 N + (5 passengers × 71 kg)) / 4
Weight per spring ≈ (13720 N + 355 N) / 4
Weight per spring ≈ 3459 N
Using the same formula as before to calculate the new frequency:
f = 1 / (2π) * √(k / m)
k = (2π * 3 Hz)² * 3459 N / kg
k ≈ 64579 N/m
Therefore, the spring constant of each spring with the presence of five passengers is approximately 64579 N/m.
f = sqrt(k/m)
where f is the frequency, k is the spring constant of the four springs, and m is the mass of the car.
=> k = f²*m = (3 Hz)²*1400 kg =12600 N/m
This results is the results for the total of the four springs, so for one spring we divide the answer by four:
k/4 = 3150 N/m
The extra weight on the car will amount to 355 kg, bringing the total of car and passengers to 1750 kg.
=> f = sqrt(k/m) = sqrt(12600 (N/m)/17150 kg) = 2.68 Hz