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A mass m is connected to two springs with equal spring constants k. In the horizontal position shown, each spring is streched by an amount [itex]\Delta a[/itex]. The mass is raised vertically and begins to oscillate up and down. Assuming that the displacement is small, and ignoring gravity, show that the motion is simple harmonic and find the period.

Here is my analysis: The mass is streched vertically such that the springs strech an amount [itex]a + \Delta a[/itex]. The restoring force, F, is acting in the direction of the springs. The horizontal component of F from both springs cancel out but the vertical component from both springs are both in the same direction. Let [itex]F \sin \theta[/itex] be the vertical component of F, [itex]\theta[/itex] being the angle the string makes with the horizontal. The sum of the vertical forces acting on m, ignoring gravity is [itex]2 F \sin \theta[/itex].

What is [itex]\sin \theta[/itex]? Some quick trig. yields:

[tex]\sin \theta = \sqrt{(a + \Delta a)^2 - a^2} / (a + \Delta a) = \sqrt{2a \Delta a + (\Delta a)^2} / (a + \Delta a)[/tex]

Since the displacement is small, it is "safe" to approximate [itex]\sin \theta[/itex] as:

[tex]\sin \theta = \sqrt{2a \Delta a} / a = \sqrt{2 \Delta a / a} [/tex]

And so the sum of the vertical forces on m is:

[tex]2 F \sin \theta = 2 (-k \Delta a) \sqrt{2 \Delta a / a} [/tex]

The above is not akin to Hooke's law so I'm wondering how in the world is the motion simple harmonic? Perhaps I did something wrong in my analysis...

Here is my analysis: The mass is streched vertically such that the springs strech an amount [itex]a + \Delta a[/itex]. The restoring force, F, is acting in the direction of the springs. The horizontal component of F from both springs cancel out but the vertical component from both springs are both in the same direction. Let [itex]F \sin \theta[/itex] be the vertical component of F, [itex]\theta[/itex] being the angle the string makes with the horizontal. The sum of the vertical forces acting on m, ignoring gravity is [itex]2 F \sin \theta[/itex].

What is [itex]\sin \theta[/itex]? Some quick trig. yields:

[tex]\sin \theta = \sqrt{(a + \Delta a)^2 - a^2} / (a + \Delta a) = \sqrt{2a \Delta a + (\Delta a)^2} / (a + \Delta a)[/tex]

Since the displacement is small, it is "safe" to approximate [itex]\sin \theta[/itex] as:

[tex]\sin \theta = \sqrt{2a \Delta a} / a = \sqrt{2 \Delta a / a} [/tex]

And so the sum of the vertical forces on m is:

[tex]2 F \sin \theta = 2 (-k \Delta a) \sqrt{2 \Delta a / a} [/tex]

The above is not akin to Hooke's law so I'm wondering how in the world is the motion simple harmonic? Perhaps I did something wrong in my analysis...

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