(e) time it takes the mass to get to the position x = -0.10 m after it has been released .12 X

College Physics
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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I'm having a hard time figuring out how to solve part e of this question. Detailed steps would be a tremendous help!

**Spring Oscillation Problem**

A mass of 0.12 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by:

\[ x(t) = (0.36 \, \text{m})\cos([20 \, \text{rad/s}]t) \]

Determine the following:

---

**(a) Amplitude of oscillation for the oscillating mass:**

- **Solution:** 0.36 m

---

**(b) Force constant for the spring:**

- **Solution:** 48 N/m

---

**(c) Position of the mass after it has been oscillating for one half a period:**

- **Solution:** -0.36 m

---

**(d) Position of the mass one-third of a period after it has been released:**

- **Solution:** -0.18 m

---

**(e) Time it takes the mass to get to the position \( x = -0.10 \, \text{m} \) after it has been released:**

- **Solution:** The provided answer, 0.12, is incorrect.

**Note:** How can you determine the time for the oscillating mass to get to a specific position from an expression for the position of the object at any time?

Solution must involve solving the equation \[ x(t) = (0.36 \, \text{m})\cos([20 \, \text{rad/s}]t) = -0.10 \, \text{m} \] to find the correct time.
Transcribed Image Text:**Spring Oscillation Problem** A mass of 0.12 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by: \[ x(t) = (0.36 \, \text{m})\cos([20 \, \text{rad/s}]t) \] Determine the following: --- **(a) Amplitude of oscillation for the oscillating mass:** - **Solution:** 0.36 m --- **(b) Force constant for the spring:** - **Solution:** 48 N/m --- **(c) Position of the mass after it has been oscillating for one half a period:** - **Solution:** -0.36 m --- **(d) Position of the mass one-third of a period after it has been released:** - **Solution:** -0.18 m --- **(e) Time it takes the mass to get to the position \( x = -0.10 \, \text{m} \) after it has been released:** - **Solution:** The provided answer, 0.12, is incorrect. **Note:** How can you determine the time for the oscillating mass to get to a specific position from an expression for the position of the object at any time? Solution must involve solving the equation \[ x(t) = (0.36 \, \text{m})\cos([20 \, \text{rad/s}]t) = -0.10 \, \text{m} \] to find the correct time.
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