A 1000 kg car carrying two 100 kg football players travels...

A 1000 kg car carrying two 100 kg football players travels over a bumpy "washboard" road with the bumps spaced 3.0 m apart. The driver finds that the car bounces up and down with maximum amplitude when he drives at a speed of $5.0 \mathrm{m} / \mathrm{s}$ $(\approx 11 \mathrm{mph}) .$ The car then stops and picks up three more $100 \mathrm{kg}$ passengers. By how much does the car body sag on its suspension when these three additional passengers get in?

Key Concepts

Natural Frequency of a Spring?Mass System

This concept refers to the rate at which a system oscillates when disturbed from its equilibrium position without any driving or damping forces. The natural frequency depends on the effective stiffness (spring constant) of the system and the mass attached. In problems involving vehicles and their suspensions, knowing the natural frequency is crucial because it can resonate with external periodic forces (like road bumps), leading to amplified oscillations.

Hooke’s Law and Static Equilibrium

Hooke’s Law states that the displacement (or sag) in a spring system is directly proportional to the force applied and inversely proportional to the spring constant (expressed as F = kx). In static equilibrium, the gravitational force on an added mass causes the spring to compress or sag until the spring force balances the weight. This relation is used to determine how much a system will deflect (sag) when additional loads are applied.

Determining the Effective Spring Constant from Dynamic Behavior

When a system is driven at a frequency that matches its natural frequency, resonance occurs and the amplitude reaches a maximum. By relating the driving frequency (derived from external conditions like periodic bumps) to the natural frequency of the system, one can infer the effective spring constant. This constant can then be used in static calculations, such as determining additional sag under extra load, by applying Hooke’s Law.

Transcript

00:01 Here in this given problem combined mass of the car with its passengers that will be 300 plus 1100 kilogram means this is 1400 kilograms spacing between the the bumps that will serve as wavelength of the wave of the oscillation so this wavelength lambda is equal to 5 .0 meter so using a relation among the wave velocity v and the frequency and the wavelength v is equal to f lambda frequency of oscillations f f f f is given as 6 .0 speed 6 .0 meter per second divided by 5 .0 meter means it comes out to be 1 .2 1 .2 thirds.

01:56 So angular frequency of this wave will be given by omega is equal to 2 5 f or this is 2 pi into 1 .2 radian per second so it comes out to be 2 .4 5 radian per second.

02:26 So spring factor or force constant of the springs of the springs of the car will be given by k is equal to m omega square means this is 1400 for the mass of the car into 2 .45 square and it comes out to be 8 ,064 pi square newton per meter so now using another expression for k which is obtained by the relation for the restoring force in the spring f is equal to k x where k is the sorry where x is the compression in the spring we get an expression for x means compression in the spring to be equal to x is equal to f by k so initial compression in the spring with the passenger on it x1 this will be for the force it will be m into g by k so for the mass this is 1400 kilogram into g 9 .8 divided by k which is 8 ,064 by square in meters it will be and final compression when the passengers get out of this car then the compression will be only because of the only under the mass of the car eleven hundred kilogram which is into 9 .8 divided by same spring factor which is 8 ,064 pi square meter so when the passengers get out of the car, the car rises up on the spring delta x is equal to x1 minus x2 means this is 1400 into 1...

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