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zaharov [31]
4 years ago
9

You have two springs. One has a greater spring constant than theother. You also have two objects, one with a greater mass than t

heother. Which object should be attached to which spring, so that theresulting spring-object system has the greatest possible period ofoscillation?
The object with the smaller mass should be attached to the springwith the greater spring constant.


The object with the smaller mass should be attached to the springwith the smaller spring constant.


The object with the greater mass should be attached to the springwith the greater spring constant.


The object with the greater mass should be attached to the springwith the smaller spring constant.
Physics
1 answer:
DIA [1.3K]4 years ago
8 0

The object with the greater mass should be attached to the spring with the smaller spring constant, so that the resulting spring-object system has the greatest possible period of oscillation.

Answer: Option D

<u>Explanation: </u>

According to the simple harmonic motions, from physics, it gives a relation between deformation force and the deflection. The more deflection results in more time period of oscillation.  

                                            F = - k x

where ‘k’ is the spring constant, and ‘F’ is the deformation force.

So, deflection is directly proportionate to forces, and inversely proportionate to its spring constant. Hence, we can derive that the force must be maximum, and hence weight must be maximum, with the spring constant lesser. Then, the deflection will be high. So, time period increases.

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A 1.31 kg object is attached to a horizontal spring of force constant 2.70 N/cm and is started oscillating by pulling it 6.20 cm
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Answer:

The answer is below

Explanation:

a) The change in energy is the difference between the final energy and the initial energy.

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\Delta E=\frac{1}{2}kA_f^2 -\frac{1}{2}kA_i^2\\\\k=force\ constant=2.7\ N/cm=270\ N/m, A_f=final\ dispalacment= 3.7\ cm=0.037\ m,\\ A_i=initial \ displacement = 6.2\ cm=0.062\ m\\\\Hence:\\\\\Delta E=\frac{1}{2}(270)(0.037)^2 -\frac{1}{2}(270)(0.062)^2\\\\\Delta E=-0.334 \ J

The negative sign shows that energy is lost to the environment. Hence 0.334 J is lost to the environment.

b) According to the law of conservation of energy, energy cannot be created or destroyed but transformed from one form to another.

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