Engery Conservation in Oscillatory Motion

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SUMMARY

The discussion centers on energy conservation in oscillatory motion, specifically involving a 0.980 kg block sliding on a frictionless surface and compressing a spring with a force constant of 245 N/m. The block's initial speed is 1.32 m/s, leading to a calculated spring compression of 0.0835 m and a contact time of 0.397 s. A key point of confusion arises regarding the time the block is in contact with the spring, which is determined to be 0.0993 s, equating to one-quarter of the oscillation period rather than half, as initially assumed.

PREREQUISITES
  • Understanding of conservation of energy principles in mechanics
  • Familiarity with Hooke's Law and spring constants
  • Knowledge of simple harmonic motion (SHM) and its equations
  • Ability to perform basic kinematic calculations
NEXT STEPS
  • Study the derivation of the conservation of energy equation in oscillatory systems
  • Learn about the characteristics of simple harmonic motion (SHM) and its mathematical representation
  • Explore the relationship between spring compression and oscillation period in detail
  • Investigate graphical representations of SHM, including sine and cosine functions
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in understanding the principles of energy conservation and oscillatory motion.

Julio Cesar
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A 0.980kg block slides on a frictionless, horizontal surface with a speed of 1.32m/s. The block encounters an unstretched spring with a force constant of 245N/m.

(a)How far is the spring compressed before the block comes to rest?

(b) How long is the block in contact with the spring before it comes to rest?




Equations: Conservation of energy using a spring.

Ei=Ef

1/2mvo2 = 1/2kA2

A= vo√m/k

Question (b): T = 2∏√m/k




3. Attempt at solution shows:

A= 0.0835m

T= 0.397s

As far as the calculations, I know the amplitude as well as the period of one complete occilation. However, when the question asks me how long is the block in contact with the spring before it comes to rest...

Well, my senses tell me that it would be half a period since the spring is compressed when the block comes in contact with it and stops. (v= 0m/s)

The answer it gives me is 0.0993s...?!

Where does this come from; that's not half a period that's a quarter of a period. The answer may be staring at be straight in the face but I just don't get it. Can someone explain why the answer is 1/4 the period instead of 1/2 the period?

Thanks.

 
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Wow! I feel like an idiot now. I've just spent the last hour just looking at the diagram, wondering why and didn't even think to draw out a sine function, lol!

Thanks so much, this helps out a ton!
 
A drawing always helps.

ehild
 

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