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vova2212 [387]
3 years ago
6

When hung from an ideal spring with spring constant k = 1.5 N/m, it bounces up and down with some frequency ω, if you stop the b

ouncing and let it swing from side to side through a small angle, the frequency of the pendulum is half the bounce frequency ω/2. What is the length of the un–stretched spring l =? Note ω should not be part of your answer.
Physics
1 answer:
Umnica [9.8K]3 years ago
7 0

Answer:

L = ¼ k g / m

Explanation:

This is an interesting exercise, in the first case the spring bounces under its own weight and in the second it oscillates under its own weight.

The first case angular velocity, spring mass system is

    w₁² = k / m

The second case the angular velocity is

    w₂² = L / g

They tell us

    w₂ = ½ w₁

Let's replace and calculate

     √ (L / g) = ½ √ (k / m)

      L / g = ¼ k / m

       

      L = ¼ k g / m

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Answer:

<em>The force that would be applied on the rope just to start moving the wagon is 122 N</em>

Explanation:

Frictional force opposes motion between two surfaces in contact. It is the force that must be applied before a body starts to move. Static friction  opposes the motion of two bodies that are in contact but are not moving. The magnitude of static friction to overcome for the body to move  can be calculated using equation 1.

F = μ x mg .............................. 1

where F is the frictional force;

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from the equation we are provide with;

       μs  = 0.25

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Using equation 1

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<em>Therefore a force of 122 N must be applied to the rope just to start the wagon.</em>

6 0
3 years ago
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Answer:

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Explanation:

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