Why the time is the quarter of a period?

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The discussion centers on the oscillation of a mass-spring system, specifically addressing why the time taken to return to the equilibrium position is one-fourth of the total period (t = T/4). It explains that the motion can be divided into three key positions: the lowest point, the equilibrium point, and the highest point. The time taken to move from the lowest position to the equilibrium is equal to the time taken to return to the lowest position, thus reinforcing the T/4 relationship. A visual diagram is suggested to clarify this concept further. Understanding this relationship is crucial for solving similar problems in oscillatory motion.
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Homework Statement


When the brick with mass 3 kg is hanged in a spring, it is lengthened 25 cm. If we lengthen the spring with 15 cm more and leave it free how may times does the brick take to come back to the equilibrium position

Homework Equations


In the solution it says t=T/4

The Attempt at a Solution


I understand d the other part, but why is t=T/4 (t-time, T-period)
 
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zade70 said:

Homework Statement


When the brick with mass 3 kg is hanged in a spring, it is lengthened 25 cm. If we lengthen the spring with 15 cm more and leave it free how may times does the brick take to come back to the equilibrium position

Homework Equations


In the solution it says t=T/4

The Attempt at a Solution


I understand d the other part, but why is t=T/4 (t-time, T-period)
Consider three positions, the lowest position of the oscillations (i.e. the release position), the equilibrium position, and the highest oscillation position. If it takes time t to rise from lowest to equilibrium, how long will it take to return to the lowest position?
 
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This diagram might help?
2uppnya.gif
 
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