A mass m sits at x = 1, which is the neutral position of two springs, the first

Question

A mass m sits at x = 1, which is the neutral position of two springs, the first of which connects the mass to the point x = 2 and has spring constant k1, and the second of which connects the mass to the origin and has spring constant k2. At t = 0 the mass is pulled to x = 1/2 and released. There is no gravity and no friction and the mass moves back and forth only on the x-axis. Where is the mass at time t? Hint: use the Euler-Lagrange equation applied to the function (kinetic energy minus potential energy). Kinetic energy is the usual (1/2)*m*v^2 and potential energy is elastic potential energy only. The position function x(t) will be the solution to the differential equation. (You can solve the diffEQ via Laplace if you want).

Explanation / Answer

FOLLOW THIS The different configurations could be ("constrain" is the end where the spring is attached to a wall or anything with infinite mass. The direction of g is indicated in all the cases): Two springs in series and mass at end (horizontal displacement). constrain-spring-spring-mass VgV Two springs in series and mass in the middle (horizontal displacement). constrain-spring-mass-spring-constrain VgV Two springs in series and mass at end (vertical displacement) constrain-spring-spring-mass >g> Two springs in series and mass in the middle (vertical displacement). constrain-spring-mass-spring-constrain >g> Two springs in series and mass in the middle (VERTICAL displacement). constrain-spring-mass-spring-constrain VgV Throughout the present expln. the springs are considered massless. In case you want to discuss the case of springs with mass, send your question to Madsci. Also, before we come to the individual cases, I will start with the concept common for all cases (but the last one), though I am sure you are completely aware of it: simple harmonic motion. Skip this part if you understand SHM pretty well. The motion of mass with a massless spring attached to it is usually simple harmonic motion (SHM). It is characterized by a sinusoidal motion in time with a single resonant frequency: X = A sin(w t) X = displacement from mean position A = amplitude t = time The acceleration is found by diffrentiating the above equation wrt time: Accn = -A w^2 sin(w t) = -w^2 X Note that whenever the acceleration (and hence the restoring force) is directly proportional to the negative displacement from the neutral postion (-X), the motion would be a SHM. What this means is that the acceleration increases as displacement from the neutral positon increases. The negative sign creeps in since the acceleration is in a direction opposite to the displacement For the case of a simple spring with a mass hanging at its end this equation would transform to: Accn = F/m = -k X/m = -w^2 X where k is the Hooke's constant for the spring. Thus, w = sqrt (k/m), where "sqrt" stands for square root. Thus the period of oscillation is given by: T = (2 pi)/w = 2 pi sqrt(m/k) Now consider the individual cases: Case 1: Two horizontal springs and mass at end and horizontal displacement on mass | |^^^^^^0^^^^^^^OMO -> | The spring is given a displacement in the horizontal plane The force exerted by a spring or spring constant k when displaced by a distance "x" from its neutral position is given by: F = k x Since the two springs are massless, the force at their point of connection is the same (i.e. is exactly transferred). Thus: k1 x1 = k2 x2 = -F .............(Eq1) where x1 and x2 are the displacements of the two springs. Taking the spring constant of the equivalent spring as K_eq, the force is also given by: F = -K_eq*(x1+x2) = k2 x2 ..............(Eq2) From equations 1 and 2, 1/K_eq = (x1/(k2 x2)) + (1/k2) = (1/k1) + (1/k2) The period of oscillations for this setup would thus be: T = 2 pi sqrt(m/K_eq) with w given by: w = sqrt(K_eq/m) Case 2: Two horizontal springs and mass at the middle and horizontal displ. on mass | | |^^^^^OMO^^^^^^| | | In this case the displacement on both the springs is the same thus reducing the situation to: Restoring force F = (k1 + k2)*x Thus the period for oscillations would be: T = 2 pi sqrt(m/(k1 + k2)) with w given by: w = sqrt((k1 + k2)/m)

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