Question

4. Two objects of masses m/ and m2 are connected by a massless spring as shown in the figure below. The spring has a natural length of L and a stiffness of k. Owo a. If x is the extension of the spring by the horizontal motion of the masses, use Newton's second law to determine the equations of motion for each object. b. Combine these equations to show that the system oscillates at a frequency of - mįm2 w2 = k/u where u =; mı+m2 This is called the reduced mass of the system. c. Assume that this represents a NaCl molecule, and that this molecule has a natural frequency of 1.14x1013 Hz, determined by measuring the light absorbed by the molecule. Find the force constant binding the molecule together.

Answer 1

It is given that the extension in the spring is 'x'. We can solve the problem as follows:

omwwo 1 is the extension of by the horizontal motion the spring of the mas - X2 - the acceleration of mass bo the acceleration of Let a be a and a mass m₂ & Let x be as a function the coordinate of time of my Newton's second :. Using have: law, we ma-kx Considering we get that the equation of motion

x - x - 1 at ² ... The equation of motion for given bytom mass m, is x = x Kx 72 Similarly, for mass me, we have m₂ a - kr - my a = kx

Date: Page No Multiplying equation @ by me we get and equations m, ma = m kr - 6 . Subtracting these equations, we have mm, (a, - a da ne m₂) kx Now it we write : x = Acoswt & Bsinot, we have arags w2(A coswt +B sinut) w²mm - (my tm) . w- m.tm .t - v Roman mm2 Writing M = m m , we have a where u=m, m₂