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Sagot :
To determine the maximum acceleration of a block in simple harmonic motion, we can use the following relationship. The formula for the maximum acceleration ([tex]\(a_{\text{max}}\)[/tex]) in simple harmonic motion is given by:
[tex]\[ a_{\text{max}} = \omega^2 \cdot A \][/tex]
where:
- [tex]\(\omega\)[/tex] is the angular frequency,
- [tex]\(A\)[/tex] is the amplitude of the motion.
Given the values:
- Amplitude, [tex]\(A = 0.040\)[/tex] meters,
- Angular frequency, [tex]\(\omega = 10.0\)[/tex] radians per second,
We substitute these values into the formula:
[tex]\[ a_{\text{max}} = (10.0 \, \text{rad/s})^2 \times 0.040 \, \text{m} \][/tex]
[tex]\[ a_{\text{max}} = 100 \, \text{(rad}^2/\text{s}^2) \times 0.040 \, \text{m} \][/tex]
[tex]\[ a_{\text{max}} = 4.0 \, \text{m/s}^2 \][/tex]
Therefore, the maximum acceleration of the block during its motion is [tex]\(4.0 \, \text{m/s}^2\)[/tex].
[tex]\[ a_{\text{max}} = \omega^2 \cdot A \][/tex]
where:
- [tex]\(\omega\)[/tex] is the angular frequency,
- [tex]\(A\)[/tex] is the amplitude of the motion.
Given the values:
- Amplitude, [tex]\(A = 0.040\)[/tex] meters,
- Angular frequency, [tex]\(\omega = 10.0\)[/tex] radians per second,
We substitute these values into the formula:
[tex]\[ a_{\text{max}} = (10.0 \, \text{rad/s})^2 \times 0.040 \, \text{m} \][/tex]
[tex]\[ a_{\text{max}} = 100 \, \text{(rad}^2/\text{s}^2) \times 0.040 \, \text{m} \][/tex]
[tex]\[ a_{\text{max}} = 4.0 \, \text{m/s}^2 \][/tex]
Therefore, the maximum acceleration of the block during its motion is [tex]\(4.0 \, \text{m/s}^2\)[/tex].
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