Oxford PAT 2016, Question 17

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5 thoughts on “Oxford PAT 2016, Question 17

  1. Hi,

    I don’t really understand why Amplitude would be 10, and don’t understand how the period can be deduced.

    Can you please help me out??

    Thanks

    1. Hi, thanks for your comment. Sorry, the answer to section (c) is just full of stupid careless mistakes. I’m away at the moment, but I’ll scan in a corrected version when I get back home. Thanks for bringing it up. In the meantime, here is what I think is the right answer.

      First let’s consider how we get the correct period. From the graph we can see that the spring exerts a force of 10N when extended by 5cm. So the spring constant is 10/0.05 = 200N/m (in the original answer I carelessly treated the spring constant as 0.02). Therefore if the mass is pulled a distance x metres from the equilibrium position, the restorative force F = -200x. Because F = ma, we have -200x = 0.02 d²x/dt², therefore d²x/dt² = -40000x. The solution of this equation is x = sin(t√40000) = sin(200t). Therefore the have a frequency of 200Hz, and this is equivalent to a period of 2π/200 seconds, which equals about 0.031 seconds.

      Second, the amplitude is the distance between the peak displacement and the equilibrium position. From the question we know that the peak displacement is 5cm, because that is the distance that the mass was pulled. Therefore the amplitude is 5cm (I carelessly treated the amplitude as the distance from peak to trough in the original answer, giving me 10cm).

      1. Hi, I think you might have made a mistake. Here: -200x = 0.02 d^2x/dt^2 , therefore d^2x/dt^2 = -10000x.
        ω^2 = 10000, ω = 100 If the amplitude is 5cm then we have: The solution of this equation is x = 0.05sin(100t) and the period would be π/50 Right?
        By the way, I really appreciate all this work you have done, it is really helping me prepare for the PAT this November.

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