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4 years ago

What is the relationship between the spring constant and the period in a mass hanging on a spring oscillation and why?

Physics

2 answers:

anygoal [31]4 years ago

3 0

Explanation:

Period of a mass on a spring is:

T = 2π√(m/k)

T is inversely proportional with the square root of k. So as the spring constant increases, the period decreases.

liubo4ka [24]4 years ago

3 0

Answer:

$T\propto \dfrac{1 }{\sqrt{k}} \text{ and } k \propto \dfrac{1}{T^{2}}$

Explanation:

The period T is inversely proportional to the square root of the spring constant, and

The spring constant is inversely proportional to the square of the period.

It all comes from Hooke's Law.

We usually get the equation for the period, but we can easily rearrange it to get an equation for the spring constant.

$\begin{array}{rcl}T &= &\mathbf{2\pi \sqrt{\dfrac{m}{k}}}\\\\T^{2} &=& 4 \pi^{2} \dfrac{m}{k}\\kT^{2}& =& 4 \pi^{2}m\\k &=& \mathbf{\dfrac{4 \pi^{2}m}{T^{2}}}\end{array}$

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1 year ago

What minimum speed does a 200 g puck need to make it to the top of a frictionless ramp that is 4.1 m long and inclined at 22 ∘?

myrzilka [38]

Answer:

5.5 m/ sec

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Because the inclined surface is frictionless so we can assume that total change of energy is zero

i-e ΔE = 0

Or we can say that difference between final and initial energy is zero i-e

Ef- Ei =0

Where,

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So, PEf= 200×9.8×1.54=3018.4 j

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