A 450 kg trailer is pulled with a constant speed over a...

A 450 - kg trailer is pulled with a constant speed over a bumpy road which is approximated by a sine function having an amplitude of 50 mm and wave length of 4 m. Assume the trailer is supported by two springs each having a stiffness of 800 N/m. a) Determine the speed of the trailer for which will cause the largest vertical motion (resonance) of the trailer. b) What would be the trailer vertical motion if the speed is 20 mph, and 60 mph? c) Determine the range of speed for which the wheels lose contact with ground.

Transcript

00:01 Here, let us see what is meant by resonance.

00:04 Resonance is the phenomenon in which the vibrational frequency of the system is equal to the natural frequency of the system.

00:24 So, when the vibrational frequency equals the natural frequency, then that phenomenon is known as resonant.

00:33 And at resonance the frequency is known as resonant frequency.

00:41 Here, in this question we are given mass, that is mass of trailer and load is equal to 250 kilogram.

00:54 The spring constant is given k is equal to 10 kilo newton per meter, delta m which is amplitude is equal to 40 mm and lambda that is wavelength is equal to 5 meter.

01:10 Here in the first part the total spring constant of the system which is represented by k of t is equal to two times the spring constant k.

01:26 We already know that the value of k is 10 kilo newton per meter so 2 into 10 which is equal to 20 kilo newton per meter is the total spring constant.

01:39 Now we need to find out the natural frequency which is represented by omega n is equal to square root of k t divided by m.

01:55 Kt represents the total spring constant which is 20 kilo newton per meter which becomes 20 into 10 power 3 newton per meter divided by the mass m which is already given 250 kilogram.

02:12 Upon solving we will get a natural frequency to be root 80 radian per second.

02:20 Next we need to find out the circular frequency.

02:25 Circular frequency is represented by omega f is equal to the equation is 2 pi into v by lambda.

02:36 So this is equal to 2 into pi into v divided by here the lambda wavelength is given which is 5 meter.

02:45 So 2 pi v by 5 radian per second will be the circular frequency.

02:51 Now at resonance we know that natural frequency will be equal to the circular frequency.

03:03 Substituting the values, natural frequency is equal to root 80 and circular frequency is equal to 2 pi into v divided by 5.

03:13 Substituting for v, we get v is equal to 5 into root 80 divided by 2 pi.

03:19 Upon solving, we will get the speed at which are, which are, which, you are...

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