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Rainbow [258]
3 years ago
6

n oscillator is driven by a sinusoidal force. The frequency of the applied force A : must be less than the natural frequency of

the oscillator. B : is independent of the natural frequency of the oscillator. C : becomes the natural frequency of the oscillator. D : must be equal to the natural frequency of the oscillator. E : must be greater than the natural frequency of the oscillator
Physics
1 answer:
lilavasa [31]3 years ago
5 0

Answer:

  B : is independent of the natural frequency of the oscillator

Explanation:

You can apply any force you like to a natural oscillator. It is independent of the natural frequency of the oscillator.

The result you get will depend on how the frequency of the applied force and the natural frequency relate to each other. It will also depend on the robustness of the oscillator with respect to the applied force.

Clearly, if the force is small enough, it will have no effect on the oscillator. If it is large enough, it will overpower any motion the oscillator may attempt. For forces in the intermediate range, there will be some mix of natural oscillation and forced behavior. One may modulate the other, for example.

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Now, when the rod is in horizontal position then final total energy will be as follows.

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Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

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The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

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So, according to the law of conservation of energy

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      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

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Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

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