The springs of a 1200-kg car compress 5.1 mm when its 59-kg driver gets into the driver's seat. If the car goes over a bump, what will be the frequency of vibrations?
Simple harmonic motion
Simple harmonic motion is a type of periodic motion in which an object undergoes oscillatory motion. The restoring force exerted by the object exhibiting SHM is proportional to the displacement from the equilibrium position. The force is directed towards the mean position. We see many examples of SHM around us, common ones are the motion of a pendulum, spring and vibration of strings in musical instruments, and so on.
Simple Pendulum
A simple pendulum comprises a heavy mass (called bob) attached to one end of the weightless and flexible string.
Oscillation
In Physics, oscillation means a repetitive motion that happens in a variation with respect to time. There is usually a central value, where the object would be at rest. Additionally, there are two or more positions between which the repetitive motion takes place. In mathematics, oscillations can also be described as vibrations. The most common examples of oscillation that is seen in daily lives include the alternating current (AC) or the motion of a moving pendulum.
![**Question:**
The springs of a 1200-kg car compress 5.1 mm when its 59-kg driver gets into the driver's seat. If the car goes over a bump, what will be the frequency of vibrations?
**Detailed Explanation:**
This question is related to the harmonic motion of the car's suspension system. We need to calculate the frequency of vibrations when the car goes over a bump. To solve this, we'll use principles from dynamics and simple harmonic motion.
**Step-by-Step Solution:**
1. **Determine the spring constant, \(k\):**
- When the 59-kg driver sits in the car, the springs compress by 5.1 mm (0.0051 m).
- The force exerted by the driver is: \(F = mg\), where \(m\) is the mass of the driver and \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)).
- Therefore, \(F = 59 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 578.2 \, \text{N}\).
Using Hooke's Law \(F = kx\), where \(x\) is the compression of the spring:
\[
k = \frac{F}{x} = \frac{578.2 \, \text{N}}{0.0051 \, \text{m}} = 113370.59 \, \text{N/m}
\]
2. **Calculate the effective mass, \(m\):**
- The total mass affecting the spring system is the mass of the car plus the mass of the driver: \(M = 1200 \, \text{kg} + 59 \, \text{kg} = 1259 \, \text{kg}\).
3. **Calculate the natural frequency, \(f\):**
- The natural frequency \(f\) of a mass-spring system is given by:
\[
f = \frac{1}{2\pi} \sqrt{\frac{k}{M}}
\]
Plugging in the values:
\[
f = \frac{1}{2\pi} \sqrt{\frac{113370.59 \, \text{N/m}}{125](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F15457f53-d05d-4d51-97b5-6d1d69bfbde3%2F2660dbf6-b7b8-43c3-96b4-e1e393d3d612%2Fcj5uehh_processed.jpeg&w=3840&q=75)
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