A car travels at a constant speed over a road that contains...

A car travels at a constant speed over a road that contains a series of equally spaced bumps. The spacing between bumps is d. The mass of the car is m, and the spring constant of the car's suspension springs is k. Because of resonance, a particularly jarring ride results. Ignoring the effect of the car's shock absorbers, which is the correct expression for the car's speed v in terms of d, m, and k, as well as some numerical constants? 0₁ O V= V= d 2π d 2A m k m 100-4√ √/=/ U= m

Transcript

00:01 Hello and welcome to this video solution of numerate.

00:05 This question is based on simple harmonic motion so let us see what is asked in the question and we will solve it so the first thing it's given is the front spring of the car suspension system has a spring constant of 1 .5d0 .6 newton per meter so let me write it the k the value of k which is which is noted as a spring constant which is 1 .5 .5 .0 .6 newton per meter.

00:32 I 1 .5 into the power 6 newton per meter and it supports the mass of 215 kgis great so the spring mass system we have with a spring of constant k and mass m that mass is 215 kgues okay and the wheel of the car has a radius of 0 .4 meters so let me write capital r as radius which is 0 .4 meters now there is this condition given the car is traveling on a bumpy road on which the distance between the bumps is equal to the circumference of the wheel and due to resonance the wheel starts to vibrate strongly when the car is traveling at a certain minimum linear speed so what is a speed made to calculate so the car when moved at a certain speed v then it resonates with the frequency omega then it actually reaches this omega the resonant frequency of the body system okay so what is the resonance frequency of the spring mass system that we need to calculate first so omega is we know omega is given by root over of k ym which is equal to root over of k we know 1 .5 times 10 to the power of 6 tab by 3 to 15 and if you calculate it you will get 83 .5 radiance per second now what is this this is a natural frequency of this spring mass system now what will happen on how when this resonance will take space the car is moving along this road this direction right and we have the wheels right the wheels are also rotating at certain speed this is v and virileas wheels are also rotating let's say at omega umegra something omega car c let's say okay so now when this rotational angular velocity of the rotation of the car is equal to this natural frequency of this spring mass system then this car is going to get why then this resonance will take place it will vibrate with very higher amplitudes okay so so we can say the angular velocity of the car, wc, is also equal to omega, right? which is equal to 83 .5 radiance per second...

Question

Please give Ace some feedback