Pregunta
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A mass hanging from a spring is set in motion and its ensuing velocity is given by \( v(t)=-3 \pi \sin \pi t \) for \( t \geq 0 \). Assume that the positive direction is upward and \( s(0)=3 \). a. Determine the position function for \( t \geq 0 \). b. Graph the position function on the interval \( [0,3] \). c. At what times does the mass reach its lowest point the first three times? d. At what times does the mass reach its highest point the first three times?

Ask by Stanley Little. in the United States
Jan 23,2025

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a. The position function is \( s(t) = 3 \cos(\pi t) \). b. The graph of \( s(t) = 3 \cos(\pi t) \) on the interval \([0, 3]\) shows peaks at \( t = 0, 2 \) seconds and troughs at \( t = 1, 3 \) seconds. c. The mass reaches its lowest point at \( t = 1 \) s, \( t = 3 \) s, and \( t = 5 \) s. d. The mass reaches its highest point at \( t = 0 \) s, \( t = 2 \) s, and \( t = 4 \) s.

Solución

Let's analyze the motion of the mass-spring system step by step. ### Given: - **Velocity Function:** \( v(t) = s'(t) = -3\pi \sin(\pi t) \) for \( t \geq 0 \) - **Initial Position:** \( s(0) = 3 \) meters (assuming meters as units) ### a. **Determine the Position Function \( s(t) \) for \( t \geq 0 \).** To find the position function \( s(t) \), we integrate the velocity function: \[ s(t) = \int v(t) \, dt + C = \int -3\pi \sin(\pi t) \, dt + C \] Let's perform the integration: \[ \begin{align*} s(t) &= -3\pi \int \sin(\pi t) \, dt + C \\ &= -3\pi \left( -\frac{\cos(\pi t)}{\pi} \right) + C \quad \text{(since } \int \sin(\pi t) \, dt = -\frac{\cos(\pi t)}{\pi} \text{)} \\ &= 3 \cos(\pi t) + C \end{align*} \] Now, apply the initial condition \( s(0) = 3 \): \[ s(0) = 3 \cos(0) + C = 3(1) + C = 3 + C = 3 \implies C = 0 \] **Final Position Function:** \[ s(t) = 3 \cos(\pi t) \] ### b. **Graph the Position Function on the Interval \( [0, 3] \).** The position function is: \[ s(t) = 3 \cos(\pi t) \] **Characteristics of the Graph:** - **Amplitude:** 3 meters - **Angular Frequency:** \( \pi \) radians/second - **Period:** \( \frac{2\pi}{\pi} = 2 \) seconds - **Phase Shift:** 0 - **Vertical Shift:** 0 **Key Points:** 1. **At \( t = 0 \):** \( s(0) = 3 \cos(0) = 3 \) meters (maximum point) 2. **At \( t = 1 \):** \( s(1) = 3 \cos(\pi \times 1) = -3 \) meters (minimum point) 3. **At \( t = 2 \):** \( s(2) = 3 \cos(2\pi) = 3 \) meters (maximum point) 4. **At \( t = 3 \):** \( s(3) = 3 \cos(3\pi) = -3 \) meters (minimum point) **Graph Sketch:** ``` s(t) | 3 +-----+-----+-----+-----+-----+-----+-----+ | * * * * | / \ / \ / \ / \ 0 +---+---+-----+---+-----+---+---+-----+---+-- t | / \ / \ / \ / \ / -3 +-----+-----+-----+-----+-----+-----+-----+ ``` - Peaks at \( t = 0, 2 \) - Troughs at \( t = 1, 3 \) ### c. **At What Times Does the Mass Reach Its Lowest Point the First Three Times?** The mass reaches its **lowest point** when \( s(t) \) is minimized. Given \( s(t) = 3 \cos(\pi t) \), the minimum value occurs when \( \cos(\pi t) = -1 \). \[ \cos(\pi t) = -1 \implies \pi t = \pi + 2\pi k \quad \text{for integer } k \geq 0 \] \[ t = 1 + 2k \quad \text{for integer } k \geq 0 \] **First Three Times:** 1. **First Minimum:** \( t = 1 + 2(0) = 1 \) second 2. **Second Minimum:** \( t = 1 + 2(1) = 3 \) seconds 3. **Third Minimum:** \( t = 1 + 2(2) = 5 \) seconds **So, the mass reaches its lowest point at \( t = 1 \) s, \( t = 3 \) s, and \( t = 5 \) s.** ### d. **At What Times Does the Mass Reach Its Highest Point the First Three Times?** The mass reaches its **highest point** when \( s(t) \) is maximized. Given \( s(t) = 3 \cos(\pi t) \), the maximum value occurs when \( \cos(\pi t) = 1 \). \[ \cos(\pi t) = 1 \implies \pi t = 0 + 2\pi k \quad \text{for integer } k \geq 0 \] \[ t = 2k \quad \text{for integer } k \geq 0 \] **First Three Times:** 1. **First Maximum:** \( t = 2(0) = 0 \) seconds 2. **Second Maximum:** \( t = 2(1) = 2 \) seconds 3. **Third Maximum:** \( t = 2(2) = 4 \) seconds **So, the mass reaches its highest point at \( t = 0 \) s, \( t = 2 \) s, and \( t = 4 \) s.**

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To find the position function, we first need to integrate the velocity function. The velocity function is given by \( v(t) = -3\pi \sin(\pi t) \). ### a. Determine the position function for \( t \geq 0 \). The position function \( s(t) \) can be found by integrating the velocity function: \[ s(t) = \int v(t) \, dt = \int -3\pi \sin(\pi t) \, dt \] Using the integral of \(\sin\), we have: \[ s(t) = 3\cos(\pi t) + C \] To find the constant \( C \), we use the initial condition \( s(0) = 3 \): \[ s(0) = 3\cos(0) + C = 3 \implies C = 0 \] Thus, the position function is: \[ s(t) = 3\cos(\pi t) \] ### b. Graph the position function on the interval \( [0,3] \). The graph of \( s(t) = 3\cos(\pi t) \) will oscillate between its maximum value, 3 (when \( t = 0 \), \( 2 \), ...), and its minimum value, -3 (when \( t = 1 \), \( 3 \), ...). This function has a period of 2.5 seconds, meaning it will reach its highest and lowest points twice in the interval [0, 3]. In the first cycle from \( t = 0 \) to \( t = 2 \), the graph will start at \( s(0) = 3 \), decrease to the lowest point \( s(1) = -3 \), and return to the maximum at \( s(2) = 3 \). In this interval, you will see a complete oscillation, starting from the maximum, reaching the minimum, and coming back up. ### c. At what times does the mass reach its lowest point the first three times? To find the times \( t \) when the mass reaches its lowest point, we set \( s(t) = -3 \): \[ 3\cos(\pi t) = -3 \implies \cos(\pi t) = -1 \] This occurs when \( \pi t = (2n + 1)\pi \) for integers \( n \), hence: \[ t = 2n + 1 \quad \text{where } n = 0, 1, 2, \ldots \] The lowest point occurs at: - For \( n = 0 \): \( t = 1 \) - For \( n = 1 \): \( t = 3 \) - For \( n = 2 \): \( t = 5 \) So the first three times are \( t = 1, 3, 5 \). ### d. At what times does the mass reach its highest point the first three times? To find the times \( t \) when the mass reaches its highest point, we set \( s(t) = 3 \): \[ 3\cos(\pi t) = 3 \implies \cos(\pi t) = 1 \] This occurs when \( \pi t = 2n\pi \), leading to: \[ t = 2n \quad \text{where } n = 0, 1, 2, \ldots \] The highest point occurs at: - For \( n = 0 \): \( t = 0 \) - For \( n = 1 \): \( t = 2 \) - For \( n = 2 \): \( t = 4 \) So the first three times are \( t = 0, 2, 4 \).

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