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

AI Thread Summary
The discussion focuses on calculating oscillation parameters for a mass on a spring. The user correctly calculates the spring constant (k) as 400 N/m and attempts to find the time period (T) and maximum velocity (V_max) using the wrong mass value of 2 kg instead of 0.2 kg. The amplitude should be based on the total displacement from the new equilibrium position, which is 2 cm, rather than the initial extension of 5 cm. Feedback emphasizes the importance of using the correct units and values in calculations. Accurate calculations are crucial for determining the oscillation parameters effectively.
lemon
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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

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

The Attempt at a Solution



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

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

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.
 
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