Consider the symmetric system of masses shown below. Note that the signs are cor
ID: 2066185 • Letter: C
Question
Consider the symmetric system of masses shown below.
Note that the signs are correct: if x1 > 0, then the rst term is negative, meaning that the left spring pulls
to the left. If x2 > x1 , then the second term is positive, meaning that the center spring pulls to the right.
a) Using similar reasoning, write the equation of motion for the second mass on the right, i.e. complete the RHS of the following equation:
b) First normal mode : Let x2 = x1 at all times. Show that equations (1) and (2) reduce, respectively, to equations of simple harmonic motion for x1 and x2 having the same frequency of oscillation. What is this frequency of oscillation 1 , in terms of k and m?
c) Sketch the positions of the two oscillating masses for this rst normal mode of oscillation (showing the springs too), at the instant when the mass on the right is at its maximum positive displacement.
d) Second normal mode : Let x2 = -x1 at all times. Show that equations (1) and (2) reduce, respectively, to equations of simple harmonic motion for x1 and x2 having the same frequency of oscillation (but different from the frequency in part (b).) What is the frequency of oscillation w2, in terms of k and m?
e) Sketch the positions of the two oscillating masses for this second normal mode of oscillation (showing the springs too), at the instant when the mass on the right is at its maximum positive displacement.
Consider the symmetric system of masses shown below. We will use Newton's law to analyze the motion of the two masses. Let the masses be displaced respectively by x1 and x2 from their equilibrium positions. First consider the mass on the left, which is acted upon by two springs. The leftmost spring is stretched by a total of x1 and thus exerts a force of - kx1 . The center spring is stretched by a total of (x2 - x1 ) and thus exerts a force of k(x2 - x1 ). Newton's law for the left mass is therefore Note that the signs are correct: if x1 > 0, then the ?rst term is negative, meaning that the left spring pulls to the left. If x2 > x1 , then the second term is positive, meaning that the center spring pulls to the right. a) Using similar reasoning, write the equation of motion for the second mass on the right, i.e. complete the RHS of the following equation: b) First normal mode : Let x2 = x1 at all times. Show that equations (1) and (2) reduce, respectively, to equations of simple harmonic motion for x1 and x2 having the same frequency of oscillation. What is this frequency of oscillation omega1 , in terms of k and m? c) Sketch the positions of the two oscillating masses for this ?rst normal mode of oscillation (showing the springs too), at the instant when the mass on the right is at its maximum positive displacement. d) Second normal mode : Let x2 = - x1 at all times. Show that equations (1) and (2) reduce, respectively, to equations of simple harmonic motion for x1 and x2 having the same frequency of oscillation (but different from the frequency in part (b).) What is the frequency of oscillation w2, in terms of k and m? e) Sketch the positions of the two oscillating masses for this second normal mode of oscillation (showing the springs too), at the instant when the mass on the right is at its maximum positive displacement.Explanation / Answer
(a)
ma=-kx2+k(x2-x1)
(b)
the (x2-x1) terms will all be zero giving
ma=-kx1 (1)
ma=-kx2 (2)
=(k/m) for both
(c)
They will both be there maximum displacement of x=x1=x2 , so the spring in the middle is unchanged, the spring on the right is compressed x and the spring on the left is stretched x.
(d)
the (x2-x1) terms will all be -2x2 =-2x1 giving
ma=-3kx1 (1)
ma=-3kx2 (2)
=(3k/m)
(e)
The right mass will be maximally displaced, implying the right spring is compressed fully,x.
The less mass will be minimally displaced, implying the left spring is compressed fully, x.
This means the middle spring will be stretched maximally, 2x.
hope that helps