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Chapter 10: Problem 70

Four people sit in a car. The masses of the people are $45 \mathrm{kg}, 52 \mathrm{kg}, 67 \mathrm{kg},\( and \)61 \mathrm{kg} .\( The car's mass is \)1020 \mathrm{kg}$. When the car drives over a bump, its springs cause an oscillation with a frequency of \(2.00 \mathrm{Hz}\). What would the frequency be if only the \(45-\mathrm{kg}\) person were present?

Short Answer

Expert verified
Answer: The frequency of oscillation with only the 45 kg person in the car would be approximately \(2.15\) Hz.

Step by step solution

01

Find the total mass of the system with all passengers

Firstly, calculate the total mass of the car and passengers. This value will be useful in finding the force constant of the springs (k). \(m_{total} = m_{car} + m_{1} + m_{2} + m_{3} + m_{4}\) \(m_{total} = 1020 kg + 45 kg + 52 kg + 67 kg + 61 kg\) \(m_{total} = 1245 kg\)
02

Calculate the angular frequency of oscillation

The frequency is given in Hz. We need to find the equivalent angular frequency, denoted by \(\omega\), using the formula \(\omega = 2 \pi f\). \(\omega_1 = 2 \pi f_1\) \(\omega_1 = 2 \pi (2 Hz)\) \(\omega_1 = 4 \pi\) rad/s
03

Determine the force constant of the springs (k)

We'll use the formula for the angular frequency of a spring mass system: \(\omega = \sqrt{\frac{k}{m}}\). We'll rearrange the equation to find k and then substitute the values of the total mass and angular frequency found in steps 1 and 2. \(k = m_{total} \omega_1^2\) \(k = 1245 kg \times (4 \pi)^2\) \(k \approx 193364.51 N/m\)
04

Find frequency with only the 45 kg person

Now let's find the new frequency when only the 45 kg person is present. We will use the formula for the angular frequency of a spring mass system: \(\omega_2 = \sqrt{\frac{k}{m_{45 + car}}}\), where \(m_{45 + car} = m_{car} + m_{45}\). \(m_{45 + car} = 1020 kg + 45 kg = 1065 kg\) \(\omega_2 = \sqrt{\frac{193364.51 N/m}{1065 kg}}\) \(\omega_2 \approx 4.30\) rad/s Now, we need to convert this angular frequency back to a regular frequency (in Hz). We can do this using the formula \(f = \frac{\omega}{2 \pi}\). \(f_2 = \frac{4.30}{2 \pi}\) \(f_2 \approx 2.15\) Hz The frequency of oscillation with only the 45 kg person in the car would be approximately \(2.15\) Hz.

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