Consider two particles of masses m1 and m2 joined to each other and to two fixed

Question

Consider two particles of masses m1 and m2 joined to each other and to two fixed walls at both ends by three identical model springs of stiffness k and natural length l0. The matrix equation of motion for the mechanical system is shown below. For certain choices of m1, m2 and k, a normal mode of the system is given by (x1(t), x2(t)) T where a: b, omega and are some constants. If a = -8.61, b = 5.88: determine how far (in cm) to the left of its equilibrium position does m1 have be initially displaced if is m2 initially 16.23 cm to the right of its equilibrium position, in order for the system to oscillate as a normal mode. Give your answer correct to 3 decimal places.

Explanation / Answer

basically, we have two equations

x1(t) = a cos(wt+) and x2(t) = bcos(wt+)

given that at t=0, x2(0) = 0.1623 we have to find x1(0)

let us assume x1(0) = z

z = -8.61cos( w*0+) ;

so z = -8.61cos --------------------------equation 1

0.1623 =5.88*cos(w*0+)

==> 0.1623 = 5.88 cos

cos = 0.0276

now substitute it in the equation 1,

we get

z = -8.61*0.0276 = - 0.2376m

z = -23.76 cm

-ve sign implies the displacement is in left direction..

so answer in 23.76 cm to left

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