(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A mass of 100g is supported by two identical springs of negligible mass having spring constants k= 50 dynes/cm. In the equilibrium position, the springs make an angle of 30 degree with the horizontal and are 100 cm in length. If the mass, m, is pulled down a distance of 2cm and released, find the period of the resulting oscillation.

2. Relevant equations

[itex]\sum F = ma [/itex]

[itex]F_{s} = -kx [/itex]

[itex]x = A cos(\omega t) + B sin(\omega t) [/itex]

[itex]\omega = \sqrt{k/m} [/itex]

[itex]P = 2\pi / \omega = 2\pi \sqrt{m/k} [/itex]

3. The attempt at a solution

I tried different approaches for this one. I started by taking the last formula I gave.

The y component of the restoring force is equal to half the restoring force (30 degree angle, sin(30) = 1/2.

Since we have 2 strings, I multiplied k by 2 in this equation, which gave me:

[itex]P = 2\pi \sqrt{100/2*50*cos(30)} = 2\pi \sqrt{100/50} = 4\pi [/itex]

I find this way too simple so I guess this isn't right...

Other things I tried:

I found that the diagonal "movement" in this case was 1.99 (extension = 101.01, compression = 99.02) using pythagorean theorem.

I then use the 2nd Newton law:

ma = -kx

100*a = -50x

a = -1/2x

a+1/2x = 0

[itex]x = A cos(\omega t) + B sin(\omega t) [/itex]

Initial conditions: x = 1,01 a = 0 @ t = 0

Therefore

[itex]x = 1.01 cos(\sqrt(1/2) t) [/itex]

[itex]P = 2\pi / \sqrt{1/2} = 8.886 [/itex]

Is any of this any good? I don't have the answer to this exercise so I have no clue if I'm even in the good direction...

Thanks!

Alaix

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# 2 springs @ 30 degrees supporting a mass

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