- #1
alaix
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Homework Statement
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.
Homework 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]
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