Simple harmonic motion (direction of acceleration)

AI Thread Summary
In a pendulum experiment, the equilibrium position is at 30 cm, with negative amplitude at 24.0 cm. The particle moves from 24.0 cm to 36.0 cm, and acceleration always acts towards the equilibrium position. When the particle is at 34.0 cm, it is moving away from equilibrium, indicating that the acceleration is negative. The discussion confirms that the concept of acceleration being directed towards equilibrium, regardless of the particle's position, is correct. Understanding the relationship between position, velocity, and acceleration is crucial for solving such problems.
kelvin macks
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Homework Statement


my question is on part d, (iv), i assume this is a pendulum experiment. the equlibrium position is at 30cm. then the negative amplitude should be located at 24.0cm. the particle moved from 24.0cm to 36.0cm. the acceleration is always acted towards the equlibrium position. so for d part(i) and (ii) should be positive since it's in the same direction with acceleration. But at 34.0cm, which is on the right of equlibrium position, (the particle travel form left to right), so the particles is moving away from equlibrium position, so the acceleration should be negative? am i right?
Because the motion is oppostite to the direction of acceleration (acceleration is towards the equlibrium position)..

please correct me if my concept is wrong. thank you!

Homework Equations


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You are right. The answer to (d) part 4 is wrong as indicated in your figure.
 
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One way to checit mathematically is to write your position function and take the derivative twice. Solve for your t values in the position function that give you the given platoon and given direction of velocity, then plug them into your acceleration function and you'll not only have the direction but the magnitude as well.
 
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