# Thread: Stumped on a simple harmonic motion question

1. ## Stumped 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:
$\displaystyle v_{max} = A\sqrt{\frac{k}{m}}$

2. Originally Posted by xxlvh
A proud deep-sea fisherman hangs a 65.0 kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.120 m. The fish is now pulled down 5.00 cm and released. What is the maximum velocity it will reach?

No idea where I am going wrong here. First, I solved for the spring constant:
$\displaystyle mg = k\Delta\l$
And I obtained 5314 N/m.
After that, I calculated $\displaystyle v_{max}$ by using:
$\displaystyle v_{max} = A\sqrt{\frac{k}{m}}$
With 0.17 m as the Amplitude and I got 1.54 m/s as the answer, which is incorrect. I also tried using conservation of energy formulas, only to get the same answer. Can anyone help me out??
$\displaystyle k = \frac{mg}{\Delta x} = \frac{65 \cdot 9.8}{.12} = 5308.3$ 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.

$\displaystyle v_{max} = A\omega = 0.05\sqrt{\frac{5308.3}{65}}$

3. Yiikes I need to pay closer attention. Thank you very much!