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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)

** Case 3: Two vertical springs in series and mass at end (vertical
displacement)**

____ < > < > < 0 > < > < > < MMBasically the same situation as described in with a small variation: since gravity comes to play here, the neutral position is not the unextended position of the spring. There is an intial displacement given by:

x0 = mg/K_eq

but since this being a constant does not in any way effect the proportionality constant between the acceleration and displacement, the T and w are as given by condition 1.

** Case 4: Two vertical springs in series and mass in the middle end
(vertical
displacement)**

____ < > < > < MM > < > < > < ---Again this is a case similar to case 2 with the initial displacement (at the neutral postion) given by:

x0 = mg/(k1+k2)

As in case 3, the w and T are the same as in by Case 2.

** Two springs in series and mass in the middle (VERTICAL
displacement).**

Intial: | | |^^^^^^^^^M^^^^^^| | | Displaced state: | | |# #| | # # | | # # | | # # | | # # | | # # | | M |The intial lengths of the two springs are l1 and l2 and their spring constants are k1 and k2 respectively. If their displacement at equilibrium position is x01 and x02 respectively, the equations

Considering vertical equilirbrium:

mg = k1 x01 cos(th01)+ k2 x02 cos(th02)

where th01 and th02 are the angles made by the springs with the vertical at equilibrium. Also,

sin(th01) = l1/(l1 + x01) and

sin(th02) = l2/(l2 + x02)

Considering horizontal equilibrium:

k1 x01 sin(th01) = k2 x02 sin(th02)

Thus we have four equations and four variables (x01, x02, k1, k2) which can be solved. Now if a slight vertical displacement is given to the mass it will start oscillating in both the x and y dimensions. The equations for this are:

ma = mg - (k1 x1 cos(th1)+ k2 x2 cos(th2))

=> m a_y = mg - (k1 x1 sqrt(1 - (l1/(l1+x1))^2) + k2 x2 sqrt(1 - (l2/(l2+x2))^2)

which is

=> m a_y = mg - ((k1 x1 y)/(l1 + x1) + k2 x2 y/(l2 + x2))

where y is the vertical displacement of the mass at any time.

Similarly, for the horizontal case too it turns out to be a non shm.

If you want to solve nice problems on SHM I recommend I.E.Irodov - I
remember
enjoying the large set of problems given there. For the case of springs
with
mass, which is a rather interesting case, do ask me via madsci.

--Arjun Kakkar

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