How Is Maximum Acceleration Calculated in a Harmonic Oscillator?

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The maximum acceleration of a harmonic oscillator can be calculated using the formula \( a_{max} = \frac{k}{m} A \), where \( k \) is the spring constant, \( A \) is the amplitude, and \( m \) is the mass of the oscillator. In this discussion, the spring constant \( k \) is confirmed to be 103 N/m, and the amplitude \( A \) is 10-6 m. The potential energy stored at time t0 = -0.5 s is given as 1 mJ, but the maximum potential energy calculated using the formula \( PE_{max} = \frac{1}{2}k A^2 \) results in approximately 5 x 10-10 Joules, indicating a discrepancy in the values provided.

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I need help with this question.

The potential energy stored in a harmonic oscillator at time t0 = -0.5 s is 1 mJ. The
spring-constant associated with the oscillator has the value k = 103 N m-1 and the
oscillation amplitude is A = 10-6 m.

Calculate the magnitude of the maximum acceleration.
 
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Is that k = 103 Nm-1 or k = 103 nm-1 ?

Actually, neither make much sense with an amplitude of only 10-6m and given the figure for energy stored. The maximum stored potential energy should be

[tex]PE_{max} = \frac{1}{2}k A^2[/tex]

which is about 5 x 10-10 Joules if k = 103 N/m.
 
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