$\displaystyle mg = k\Delta\l$

And I obtained 5314 N/m.

After that, I calculated $\displaystyle v_{max}$ by using:

$\displaystyle v_{max} = A\sqrt{\frac{k}{m}} $

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- Sep 21st 2009, 04:35 PMxxlvhStumped on a simple harmonic motion question
$\displaystyle mg = k\Delta\l$

And I obtained 5314 N/m.

After that, I calculated $\displaystyle v_{max}$ by using:

$\displaystyle v_{max} = A\sqrt{\frac{k}{m}} $ - Sep 21st 2009, 05:05 PMskeeter
$\displaystyle k = \frac{mg}{\Delta x} = \frac{65 \cdot 9.8}{.12} = 5308.3$ N/m

the fish is pulled down 5.0 cm = 0.05 m down from equilibrium ...

$\displaystyle A = 0.05$ m

amplitude is measured from equilibrium, not the unstretched position of the spring.

$\displaystyle v_{max} = A\omega = 0.05\sqrt{\frac{5308.3}{65}}$ - Sep 21st 2009, 05:45 PMxxlvh
Yiikes I need to pay closer attention. Thank you very much!