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WITCHER [35]
3 years ago
4

Imagine a mass hanging from a spring that behaves ideally. Which factor will not affect the period of this mass-spring system?

Physics
2 answers:
natulia [17]3 years ago
0 0

Answer:

varying the displacement of the mass from its equilibrium position

increasing the amount of time the oscillating system is observed

keeping track of the precise position of the mass through time

Explanation:

The period of a mass-spring system is given by:

T=2 \pi \sqrt{\frac{m}{k}}

where

m is the mass hanging from the spring

k is the spring constant

As we can see from the formula, the period depends only on these two factors: mass and spring constant, and nothing else. Therefore, the following factors do not affect the period of the system:

varying the displacement of the mass from its equilibrium position

increasing the amount of time the oscillating system is observed

keeping track of the precise position of the mass through time

insens350 [35]3 years ago
0 0
T=2\pi \sqrt{\frac{m}{k} }

Where T is the period, m is the mass of the object, and k is the spring constant. It is important to note that the amplitude of the oscillation has no effect on the period. 
Which means the question should in fact say factors - the displacement will have no effect, nor will observing it in any way - the only two things there that will affect the period of the spring is changing the weight and the spring constant.
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0 = (MR²/2)(ω²) - mv²r ................where is ω is angular speed which is required in part (a) of question

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