The body of a 1269 kg car is supported on a
frame by four springs. The spring constant of
a single spring is 2.75 × 10
4
N/m. Four people riding in the car have a combined mass
of 256 kg. When driven over a pothole in
the road, the frame vibrates and for the first
few seconds the vibration approximates simple harmonic motion.
What is the period of vibration of the car?
when springs are in parallel, the following happens:
F/4=kx
k=F/4x which is exactly 1/4 of the original k.
so here the combined spring constant then is 2.75E4 /4
mass total...you do it.
F= 1/2PI sqrt (k/m)
and
t=1/F= 2PI sqrt (m/k)
To find the period of vibration of the car, we need to use the formula:
T = 2π √(m/k)
where T is the period of vibration, π is a mathematical constant (approximately equal to 3.14), m is the total mass of the system (car + passengers), and k is the spring constant.
In this case, the total mass (m) is the mass of the car (1269 kg) plus the mass of the passengers (256 kg), which gives us a total mass of 1525 kg.
The spring constant (k) is given as 2.75 × 10^4 N/m.
Now, we can substitute the values into the formula and calculate the period of vibration:
T = 2π √(m/k)
T = 2π √(1525 kg / 2.75 × 10^4 N/m)
Calculating this expression gives us the period of vibration of the car.