Lesson 7, Topic 3
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# Horizontal Oscillations of a Spring-Mass System

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Consider a system containing a spring with a block of mass m attached to it on a smooth horizontal surface (frictionless surface) as shown below. The spring constant = k.

At this point, the block is at its central or equilibrium position, O.

If the mass is pulled towards the right, through a small displacement x from its equilibrium position, and then released, it will oscillate back and forth its equilibrium position, O.

Once released, it will move back towards its equilibrium position, O, and overshoot it to a new position, Y which is -x from O.

From Y, its compressed position, there will be a restoring force that will take it past O and back to A.

Consider the expansion of the spring;

Let  be the restoring force that tries to bring the mass back from Z to the equilibrium position O.

The force will be negative because it acts opposite to the direction of the displacement.

F = -kx

As the mass is released at Z it will have an acceleration;

a = $$\frac{F}{m}$$

where F is the restoring force

and m is the mass of the block

∴ a = $$\frac{-k}{m} \scriptsize x$$

for a given spring system $$\frac{-k}{m}$$ is constant

∴ $$\scriptsize a \propto -x$$

This is the condition for S.H.M. Therefore when the mass is oscillating around the mean position it is performing simple harmonic motion.

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