Oscillation of object on spring - Find speed

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SUMMARY

The discussion centers on calculating the speed of an object oscillating on a massless spring, specifically when it is 8 cm below its initial position. The initial claim states the speed is 56 cm/s, but the correct calculation, based on energy conservation principles, reveals that the speed is actually 42 cm/s when the object is 3 cm below the equilibrium point. The angular velocity is determined to be 14 rad/s, but the method of using energy conservation is recommended for accuracy over angular displacement calculations.

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  • Understanding of Hooke's Law and spring constants
  • Basic principles of oscillatory motion
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A massless spring hangs from the ceiling with a small object attached to its lower end. the object is initially held at rest in a position y such that the spring is at its rest length(not stretched). The object is then released from y and oscillates up and down, with its lowest position being 10 cm below y. What is the speed of the object when it is 8cm below the initial position?

The book says 56 cm/s, but I don't agree with it. First of all, the amplitude will be 5 cm, which, we can use to figure out the angular velocity.

K(.05) = Fg = mg

K = 196m

angular velocity = sqrt(K/m)
= 14 rad/s.

So when the object reaches the point, 8cm from y, it would've displaced only 3 cm from the "equilibrium point".

so V = angular velocity * displacement
= 14 * 3 = 42 cm/s

Am I correct here? Or is there something I'm missing? Thanks.
 
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Yes, you are missing something. The object is not undergoing simple circular motion.

Even though you can solve for the correct motion I suggest using energy conservation instead.
 

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