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Four passengers with combined mass 250 kg compress the springs of a car with worn-out shock absorbers by 4.00 cm when they get in. Model the car and passengers as a single body on a single ideal spring. If the loaded car has a period of vibration of 1.92 s, what is the period of vibration of the empty car?

Short Answer

Expert verified
The period of vibration of the empty car is 1.72 s.

Step by step solution

01

Understanding the System

Given that the car and passengers are modeled as a body on a spring, the spring constant can be determined using Hooke's Law. We know that the combined weight of passengers (250 kg) compresses the spring by 4.00 cm (0.04 m).
02

Calculate the Spring Constant

Use Hooke's Law given by \[ F = kx \]where \( F \) is the force due to weight, \( k \) is the spring constant, and \( x \) is the displacement. Here, \( F = mg = 250 \times 9.8 \) N, and \( x = 0.04 \) m. Solve for \( k \):\[ k = \frac{250 \times 9.8}{0.04} = 61250 \, \text{N/m} \]
03

Determine the Loaded Period

We are given that the loaded car (car + passengers) has a period of 1.92 s. The period \( T \) of a mass-spring system is given by:\[ T = 2\pi\sqrt{\frac{m}{k}} \]Using this, rearrange to find \( m \) (loaded mass):\[ 1.92 = 2\pi\sqrt{\frac{m}{61250}} \]Solve for \( m \).
04

Find Total Loaded Mass

Solve \( 1.92 = 2\pi\sqrt{\frac{m}{61250}} \).\[ \frac{1.92}{2\pi} = \sqrt{\frac{m}{61250}} \]\[ \left(\frac{1.92}{2\pi}\right)^2 \times 61250 = m \]The loaded mass \( m \) includes car plus passengers.
05

Calculate Empty Car Mass

As the passengers have a combined mass of 250 kg:\[ m_\text{car alone} = m - 250 \]
06

Determine Empty Car Period

Now, use the empty car mass to find its period \( T_\text{empty} \):\[ T_\text{empty} = 2\pi\sqrt{\frac{m_\text{car alone}}{61250}} \]Calculate for \( T_\text{empty} \).
07

Calculation Summary

Upon calculation, using the determined empty car mass, we find:\[ T_\text{empty} = 1.72 \, ext{s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spring Constant
The spring constant is a measure of a spring's stiffness. It tells us how much force is needed to compress or stretch the spring by a unit length. When we think of a car with worn-out shock absorbers, the spring constant determines how much the car will lower when passengers get in.
In our exercise, we found the spring constant using the formula from Hooke’s Law:
  • Force (\( F = mg \)) involves weight, which is mass times gravitational acceleration.
  • The displacement (\( x \) is the compression of the spring.
This allowed us to calculate the spring constant (\( k \) as 61250 N/m, indicating the car's springs are quite stiff.
Hooke's Law
Hooke's Law explains the behavior of springs in linear terms. It shows that the displacement of the spring is directly proportional to the force applied. The law's formula is\[ F = kx \]where:
  • \( F \) is the force applied in Newtons.
  • \( k \) is the spring constant in N/m.
  • \( x \) is the displacement in meters.
In this context, when 250 kg of passengers enter the car, they cause a displacement of 0.04 m. This displacement is due to the force exerted by the passengers' weight, allowing us to use Hooke's Law to find how strong the spring is.
Period of Vibration
The period of vibration refers to how long it takes for one complete cycle of motion in a simple harmonic oscillator, like a mass-spring system. It depends on both the mass and the spring constant.
The formula to find the period (\( T \) is given by:\[ T = 2\pi\sqrt{\frac{m}{k}} \]We use this to calculate the period of the loaded car and then the empty car.
For our loaded car, the period was given as 1.92 seconds. This information helps calculate the total mass involved and later determine the empty car's period of vibration.
Mass-Spring System
The concept of a mass-spring system is a fundamental example of harmonic motion. In this system, the mass oscillates back and forth after being displaced from its equilibrium position. This setup is beautifully demonstrated in the car model with passengers and the spring acting as the shock absorber.
In our problem:
  • The mass is the combined weight of the car and the passengers.
  • The spring is modeled by the car's suspension.
By understanding how these components interact, we can analyze how the system's period changes when mass (passengers) is added or removed. This gives us a clear picture of how modifications, like adding weight, affect the dynamics of the mass-spring system.

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Most popular questions from this chapter

An object is undergoing \(\textbf{SHM}\) with period 0.900 s and amplitude 0.320 m. At \(t =\) 0 the object is at \(x =\) 0.320 m and is instantaneously at rest. Calculate the time it takes the object to go (a) from \(x =\) 0.320 m to \(x =\) 0.160 m and (b) from \(x =\) 0.160 m to \(x =\) 0.

While on a visit to Minnesota ("Land of 10,000 Lakes"), you sign up to take an excursion around one of the larger lakes. When you go to the dock where the 1500-kg boat is tied, you find that the boat is bobbing up and down in the waves, executing simple harmonic motion with amplitude 20 cm. The boat takes 3.5 s to make one complete up-and-down cycle. When the boat is at its highest point, its deck is at the same height as the stationary dock. As you watch the boat bob up and down, you (mass 60 kg) begin to feel a bit woozy, due in part to the previous night's dinner of lutefisk. As a result, you refuse to board the boat unless the level of the boat's deck is within 10 cm of the dock level. How much time do you have to board the boat comfortably during each cycle of up-and-down motion?

A small sphere with mass \(m\) is attached to a massless rod of length \(L\) that is pivoted at the top, forming a simple pendulum. The pendulum is pulled to one side so that the rod is at an angle \(\theta\) from the vertical, and released from rest. (a) In a diagram, show the pendulum just after it is released. Draw vectors representing the \(forces\) acting on the small sphere and the \(acceleration\) of the sphere. Accuracy counts! At this point, what is the linear acceleration of the sphere? (b) Repeat part (a) for the instant when the pendulum rod is at an angle \(\theta\)/2 from the vertical. (c) Repeat part (a) for the instant when the pendulum rod is vertical. At this point, what is the linear speed of the sphere?

A thrill-seeking cat with mass 4.00 kg is attached by a harness to an ideal spring of negligible mass and oscillates vertically in SHM. The amplitude is 0.050 m, and at the highest point of the motion the spring has its natural unstretched length. Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring), the kinetic energy of the cat, the gravitational potential energy of the system relative to the lowest point of the motion, and the sum of these three energies when the cat is (a) at its highest point; (b) at its lowest point; (c) at its equilibrium position.

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude of the motion is 0.250 m and the period is 3.20 s. What are the speed and acceleration of the block when \(x =\) 0.160 m?

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