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A small trailer and its load have a total mass of $250 \mathrm{kg}$. The trailer is supported by two springs, and when the boat is given an initial displacement $x_{0}$ downward and released from rest, the amplitude of oscillation is reduced to $1 / 4 x_{0}$ after one half cycle and the period of the oscillation is 0.8 s. The trailer is then pulled over a road, the surface of which can be approximated by a sine curve with an amplitude of $35 \mathrm{mm}$ and a wavelength of $5 \mathrm{m}$ (i.e., the distance between successive crests is $5 \mathrm{m}$ and the vertical distance from crest to trough is $70 \mathrm{mm}$ ). Determine $(a)$ the damping ratio, $(b)$ the spring stiffness, $(c)$ the amplitude of the vibration of the trailer at a speed of $50 \mathrm{km} / \mathrm{h}$. (Hint:Use the logarithmic decrement discussed in Probs. 19.129 and $19.130 .$

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Mechanical Vibrations

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Hi everyone, a son in figures, l mass m, is displaced to the position x y and the large arion of the system will be kinetic: energy, minus potential energy, half m x, dot square plus, half m y dot, squareplus half k y square plus plus x, squared Minus n, not plus half k, y square plus l minus x, whole square, minus n, not square to find equation of original motion, use the edward flaring equation. It would be d upon duty died over das is equal to a porthis, will give you m x. Double dot is called minus k, root of y square plus n plus x whole square, minus n, not into head plus x, over root of y square plus n plus x square, and this we can write minus k, root of y square plus. This is minus y square plus l minus x whole square, minus l, not into l minus x, over y square plus n minus x square minus 1. So this could be written as minus k. 1, minus n, not root of y square plus plus x whole square into l plus x, plus k 1 minus n, not over y square plus l minus x whole square into l minus x. What do you and this result could be simplified? Minus k, 2 x minus and not into 1 upon root of y square, plus l plus x, pi minus 1 upon root of y square plus l minus x, whole square. Minus n not x 1 upon y square plus l plus x, squared plus 1 divided by root of y squared plus l minus x square 4, l plus minus x, to be greater than by this to be approximated to m x. Double dot is equal to k. 2 x, minus l not 1 upon l plus x, upon l, minus x and finally, you can write m x, double da 2 v minus k, 2 x minus and not 1 over on l plus x, minus l over l minus x, minus l not x 1. Over l plus x, upon 1 minus l upon x and this to be approximated to minus 2 x. Thus the equation along x axis, you will get m x, 2 k x is equal to 0 and this is first part now. Second part g over dt of y over dell y dot dal by it will give you m y double dot is equalled to k, root of y square plus l plus x, whole square, minus l, not into y upon y square plus en plus x, whole square And this could be simplified to m y double dot, plus 2 k, 1 minus and not upon l into y is equal to 0, see put the ratio of periods equal to t x upon d yomega y omega root of 2 k, 1 minus n, not y Divided by m upon root of 2 k, i m, so it can be written as root of 1 minus n, not by a dart in the motion is start from 0 velocity from 0 velocity. Then position of mass h, described by x, is equal to a not sine of omega x and y is equal to a not cos of mega by teary t graph and x graph are given as thanks for watching it.

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