Four people sit in a car. The masses of the people are 41 kg, 47 kg, 53 kg, and 55 kg. The car's mass is 1020 kg. When the car drives over a bump, its springs cause an oscillation with a frequency of 1.00 Hz. What would the frequency be if only the 41-kg person were present?

Hz

I really don't know how to do this, and I can't even make an assumption! Help please!

M=41+47+53+55 = 196 kg

f=sqrt(k/M)/ 2•π
k =(2•π •f)^2•M =7.74•10^3 N/m
f1 =sqrt(k/m)/ = sqrt(7.74•10^3/41)/ 2•π =2.19 Hz.

Thank you, but I tried this answer its wrong and I don't know why!

Thanks, for the help Elena. For me this answer is also not working its saying that its wrong. :/

add the mass of the vehicle...

To find the frequency of the car's oscillation with only the 41-kg person present, you can use the principle of simple harmonic motion.

First, we need to calculate the effective mass of the system when only the 41-kg person is present. The effective mass is the sum of the masses of the car and the person. So, the effective mass would be:

Effective mass = Mass of the car + Mass of the person
= 1020 kg + 41 kg
= 1061 kg

Now, we can use the equation for the frequency of simple harmonic motion:

Frequency (f) = 1 / (2π * sqrt(m / k))

Where:
- f is the frequency in Hz
- π is a mathematical constant (approximately 3.14159)
- m is the effective mass of the system
- k is the spring constant

Since the problem doesn't provide the spring constant, we cannot calculate the exact frequency. However, we can still determine the relationship between the two frequencies.

The frequency of the car's oscillation with all four people present is given as 1.00 Hz. Let's call this frequency f1. Therefore, we have:

f1 = 1.00 Hz

Now, let's assume that the spring constant remains the same when only the 41-kg person is present. Let's calculate the frequency using the effective mass of 1061 kg. Let's call this frequency f2. Therefore, we have:

f2 = 1 / (2π * sqrt(1061 kg / k))

Since k is the same for both scenarios, we can express the relationship between the two frequencies as:

f1 / f2 = (1.00 Hz) / (1 / (2π * sqrt(1061 kg / k)))
= 2π * sqrt(1061 kg / k)

To find the frequency f2 when only the 41-kg person is present, we would need the value of k or some additional information about the system. Without that information, we cannot calculate the exact frequency, but we can determine that it would be less than the original frequency of 1.00 Hz.