How Do You Calculate Oscillation Parameters for a Mass on a Spring?

[lemon]

Hi:
Would somebody kindly check I have approached this problem correctly, please?
Thank you.

Homework Statement

A 0.2kg mass is suspended from a light spring, producing an extension of 5.0cm.

a) Calculate the force constant of the spring.

The mass is now pulled down a further 2.0cm and released.

b) Calculate
i) the time period of the resultant oscillations
ii) the maximum velocity of the mass

c) sketch a displacement time graph for one complete oscillation of the mass, marking amplitude (A) and time period (T) and showing where maximum velocity occurs

Homework Equations

$$\begin{array}{l} F = k\Delta x \\ T = 2\pi \sqrt {\frac{m}{k}} \\ f = \frac{1}{T} \\ V_{\max } = 2\pi fA \\ \end{array}$$

The Attempt at a Solution

$$\begin{array}{l} a){\rm{ }}k = \frac{{20}}{{0.05}} \\ = 400 \\ \\ b){\rm{ i) T = 2}}\pi \sqrt {\frac{2}{{400}}} \\ = 0.4442s{\rm{ }}\left( {4s.f.} \right) \\ \\ {\rm{ii) f = }}\frac{1}{{0.4442}} \\ = 2.2624Hz{\rm{ }}\left( {4s.f.} \right) \\ V_{\max } = 2\pi \times 2.2624 \times 0.07 \\ = 0.9951{\rm{ }}\left( {4s.f.} \right) \\ 1.0ms^{ - 1} {\rm{ }}\left( {2s.f.} \right) \\ \end{array}$$

I have taken the Amplitude from the spring equilibrium and not the equilibrium when the mass attached to the spring with initial extension of 5.0cm. Is this correct or should the Amplitude be just the secondary extension of 2.0cm?
Thank you

 

[ideasrule]

$$\begin{array}{l} a){\rm{ }}k = \frac{{20}}{{0.05}} \\ = 400 \\ \\ b){\rm{ i) T = 2}}\pi \sqrt {\frac{2}{{400}}} \\ = 0.4442s{\rm{ }}\left( {4s.f.} \right) \\ \\ {\rm{ii) f = }}\frac{1}{{0.4442}} \\ = 2.2624Hz{\rm{ }}\left( {4s.f.} \right) \\ V_{\max } = 2\pi \times 2.2624 \times 0.07 \\ = 0.9951{\rm{ }}\left( {4s.f.} \right) \\ 1.0ms^{ - 1} {\rm{ }}\left( {2s.f.} \right) \\ \end{array}$$

First, you need units!

Part b) isn't correct. Why did you use 2 kg for m?

Also, you need to use 2 cm, not 7 cm. This is because amplitude is always the displacement from the equilibrium point, and when the mass is attached, the equilibrium point is 2 cm away.