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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

A large, 34.0-kg bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell's center of mass is 0.60 m below the pivot. The bell's moment of inertia about an axis at the pivot is 18.0 kg \(\cdot\) m\(^2\). The clapper is a small, 1.8-kg mass attached to one end of a slender rod of length \(L\) and negligible mass. The other end of the rod is attached to the inside of the bell; the rod can swing freely about the same axis as the bell. What should be the length \(L\) of the clapper rod for the bell to ring silently\(-\)that is, for the period of oscillation for the bell to equal that of the clapper?

A holiday ornament in the shape of a hollow sphere with mass \(M =\) 0.015 kg and radius \(R =\) 0.050 m is hung from a tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and released, it swings back and forth as a physical pendulum with negligible friction. Calculate its period. (\(Hint\): Use the parallel-axis theorem to find the moment of inertia of the sphere about the pivot at the tree limb.)

In a physics lab, you attach a 0.200-kg air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is 2.60 s. Find the spring's force constant.

An object with mass 0.200 kg is acted on by an elastic restoring force with force constant 10.0 N/m. (a) Graph elastic potential energy \(U\) as a function of displacement \(x\) over a range of \(x\) from \(-\)0.300 m to \(+\)0.300 m. On your graph, let 1 cm \(=\) 0.05 J vertically and 1 cm \(=\) 0.05 m horizontally. The object is set into oscillation with an initial potential energy of 0.140 J and an initial kinetic energy of 0.060 J. Answer the following questions by referring to the graph. (b) What is the amplitude of oscillation? (c) What is the potential energy when the displacement is one-half the amplitude? (d) At what displacement are the kinetic and potential energies equal? (e) What is the value of the phase angle \(\phi\) if the initial velocity is positive and the initial displacement is negative?

A 1.50-kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 185 N/m and a 275-g metal ball is in the tray. The spring is below the tray, so it can oscillate up and down. The tray is then pushed down to point \(A\), which is 15.0 cm below the equilibrium point, and released from rest. (a) How high above point \(A\) will the tray be when the metal ball leaves the tray? (\(Hint\): This does \(not\) occur when the ball and tray reach their maximum speeds.) (b) How much time elapses between releasing the system at point \(A\) and the ball leaving the tray? (c) How fast is the ball moving just as it leaves the tray?

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