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Sagot :
To find the times when the weight is halfway to its maximum negative position, we start by identifying the necessary components of a sinusoidal function that describes the motion of the weight.
### Step-by-Step Solution:
1. Identify the Equation of Motion:
- A general sinusoidal function can be written as [tex]\( y(t) = A \sin(Bt + C) \)[/tex], where:
- [tex]\( A \)[/tex] is the amplitude (maximum displacement),
- [tex]\( B \)[/tex] is the angular frequency,
- [tex]\( C \)[/tex] is the phase shift.
2. Determine the Specific Values:
- In this case, let's assume the amplitude [tex]\( A \)[/tex] is 1 (you may adjust if a different amplitude is defined).
- We know that the weight is halfway to its maximum negative position, so [tex]\( y(t) = -\frac{A}{2} \)[/tex], which simplifies to [tex]\( y(t) = -0.5 \)[/tex].
3. Set Up the Equation:
- The equation becomes [tex]\( \sin(Bt + C) = -0.5 \)[/tex].
4. Solve for [tex]\( t \)[/tex]:
- To solve this equation, we utilize the inverse sine function [tex]\( \arcsin \)[/tex].
- Let's assume [tex]\( B = 2\pi \)[/tex] (indicating one complete oscillation per second) and [tex]\( C = 0 \)[/tex] (assuming no phase shift).
Thus, we get:
[tex]\[ \sin(2\pi t) = -0.5 \][/tex]
- Solving [tex]\( \sin(2\pi t) = -0.5 \)[/tex]:
[tex]\[ 2\pi t = \arcsin(-0.5) \][/tex]
[tex]\[ t_1 = \frac{\arcsin(-0.5)}{2\pi} \][/tex]
5. Calculate [tex]\( t_1 \)[/tex]:
[tex]\[ \arcsin(-0.5) = -\frac{\pi}{6} \][/tex]
[tex]\[ t_1 = \frac{-\frac{\pi}{6}}{2\pi} = -\frac{1}{12} \approx -0.08 \][/tex]
6. Since sine is periodic, find the next time [tex]\( t_2 \)[/tex]:
[tex]\[ 2\pi t = \pi - \arcsin(-0.5) \][/tex]
[tex]\[ 2\pi t = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \][/tex]
[tex]\[ t_2 = \frac{\frac{7\pi}{6}}{2\pi} = \frac{7}{12} \approx 0.58 \][/tex]
7. Check the interval [tex]\( 0 \leq t \leq 0.5 \)[/tex]:
- Neither [tex]\( t_1 \approx -0.08 \)[/tex] nor [tex]\( t_2 \approx 0.58 \)[/tex] falls within the interval [tex]\([0, 0.5]\)[/tex].
### Final Answer:
The times [tex]\( t \)[/tex] when the weight is halfway to its maximum negative position in the interval [tex]\( 0 \leq t \leq 0.5 \)[/tex] are:
- None of these times fall in the given interval.
Thus, there are no times within the interval [tex]\( 0 \leq t \leq 0.5 \)[/tex] when the weight is halfway to its maximum negative position.
### Step-by-Step Solution:
1. Identify the Equation of Motion:
- A general sinusoidal function can be written as [tex]\( y(t) = A \sin(Bt + C) \)[/tex], where:
- [tex]\( A \)[/tex] is the amplitude (maximum displacement),
- [tex]\( B \)[/tex] is the angular frequency,
- [tex]\( C \)[/tex] is the phase shift.
2. Determine the Specific Values:
- In this case, let's assume the amplitude [tex]\( A \)[/tex] is 1 (you may adjust if a different amplitude is defined).
- We know that the weight is halfway to its maximum negative position, so [tex]\( y(t) = -\frac{A}{2} \)[/tex], which simplifies to [tex]\( y(t) = -0.5 \)[/tex].
3. Set Up the Equation:
- The equation becomes [tex]\( \sin(Bt + C) = -0.5 \)[/tex].
4. Solve for [tex]\( t \)[/tex]:
- To solve this equation, we utilize the inverse sine function [tex]\( \arcsin \)[/tex].
- Let's assume [tex]\( B = 2\pi \)[/tex] (indicating one complete oscillation per second) and [tex]\( C = 0 \)[/tex] (assuming no phase shift).
Thus, we get:
[tex]\[ \sin(2\pi t) = -0.5 \][/tex]
- Solving [tex]\( \sin(2\pi t) = -0.5 \)[/tex]:
[tex]\[ 2\pi t = \arcsin(-0.5) \][/tex]
[tex]\[ t_1 = \frac{\arcsin(-0.5)}{2\pi} \][/tex]
5. Calculate [tex]\( t_1 \)[/tex]:
[tex]\[ \arcsin(-0.5) = -\frac{\pi}{6} \][/tex]
[tex]\[ t_1 = \frac{-\frac{\pi}{6}}{2\pi} = -\frac{1}{12} \approx -0.08 \][/tex]
6. Since sine is periodic, find the next time [tex]\( t_2 \)[/tex]:
[tex]\[ 2\pi t = \pi - \arcsin(-0.5) \][/tex]
[tex]\[ 2\pi t = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \][/tex]
[tex]\[ t_2 = \frac{\frac{7\pi}{6}}{2\pi} = \frac{7}{12} \approx 0.58 \][/tex]
7. Check the interval [tex]\( 0 \leq t \leq 0.5 \)[/tex]:
- Neither [tex]\( t_1 \approx -0.08 \)[/tex] nor [tex]\( t_2 \approx 0.58 \)[/tex] falls within the interval [tex]\([0, 0.5]\)[/tex].
### Final Answer:
The times [tex]\( t \)[/tex] when the weight is halfway to its maximum negative position in the interval [tex]\( 0 \leq t \leq 0.5 \)[/tex] are:
- None of these times fall in the given interval.
Thus, there are no times within the interval [tex]\( 0 \leq t \leq 0.5 \)[/tex] when the weight is halfway to its maximum negative position.
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