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Problem 4: An object is attached to a spring in such a way that it undergoes...

Question

Problem 4: An object is attached to a spring in such a way that it undergoes...

Problem 4: An object is attached to a spring in such a way that it undergoes simple harmonic motion with a period of 2 s. a)
d) [4 points] Under the assumption above, what is the maximum velocity and acceleration of the object? e) [5 points] Write do
f) *BONUS* [3 points) Write down the formula for the position of the object as a function of time, x(t), assuming that its ve
Problem 4: An object is attached to a spring in such a way that it undergoes simple harmonic motion with a period of 2 s. a) [3 points] What is the angular frequency of the oscillations? b) [3 points] If the mass of the object is 4 kg, what is the value of the spring constant, k? c) [2 points] If the amplitude of the oscillations is 20 cm, what is the total energy of the spring- mass system?
d) [4 points] Under the assumption above, what is the maximum velocity and acceleration of the object? e) [5 points] Write down the formula for the position of the object as a function of time, x(t), assuming that its velocity is maximum at t = 0 going toward the positive direction. Identify the value of all the constants that you use.
f) *BONUS* [3 points) Write down the formula for the position of the object as a function of time, x(t), assuming that its velocity is maximum at t = 0 going toward the positive direction. Identify the value of all the constants that you use. (If you use sine version for part (e), then you need to use the cosine version here, and vice versa)
Science Physics
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Answers

(a) So for a simple harmonic motion the angular frequency is given by, ω = 2πν 2π T Here T = 2 s. So \omega = \pi ~ Hz.

(b)\omega = \sqrt{\frac{k}{m}}. Here, m = 4 ~ kg. So, k = m \omega^2 = 4 \cdot \pi ^2~ N/m = 4 \pi^2 ~ N/m.

(c) E_{\text{max}} = \frac12 k x_{\text{max}}^2. Here, x_{\text{max}} = 20.0 ~ cm = 20.0 \times 10^{-2} ~ m. So, E = \frac12 \cdot 4\pi^2 \cdot 20.0 \times 10^{-2} ~ J = \frac25 \pi^2 ~ J.

(d) Using conservation of energy, E_{\text{max}} = \frac12 m v_{\text{max}}^2. So, v_{\text{max}} = \sqrt{\frac{2 E_{\text{max}}}{m}} = \sqrt{\frac45} \pi ~ m/s.

Now for a simple harmonic motion one already knows (from the equation of motion),F = -k x \implies a = \frac{k x}{m} \implies a_{\text{max}} = \frac{k}{m} x_{\text{max}}.

So, a_{\text{max}} = a_{\text{max}} = \frac{\pi^2}{5} ~ m/s^2.

(e) As already mentioned in the previous subsection, the equation of motion is ,F = -k \cdot x \implies m \ddot{x} = - k \cdot x. Solving this one gets (with the initial conditions, x(0) = 20.0 ~ cm. \dot{x} = 0.), x(t) = 20.0 \cos(\pi \cdot t) ~ m.

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