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    Hi guys, I'm doing A2 Physics for OCR, and I've encountered a past paper question about a trolley attached to a spring (which is attached to a vertical unmoving surface) on a smooth table that follows simple harmonic motion, which is as follows;

    'Using your knowledge of Hooke's Law and Newton's second law, determine the period T of the trolley in terms of the force constant k of the spring and the mass m of the trolley.'

    (w = omega)

    Now, I did F = kx = ma; I then put a = w^2x into it, to obtain kx = mw^2x; I cancelled the x on both sides, and then divided by m to obtain w^2 = k/m, as in the answers.

    I then square rooted both sides, flipped both sides over so that T was on top for w, and then multiplied through so that T = 2pi(sqrt(m/k)).

    The answer, however, says that I should have got T = 2pi(sqrt(k/m)), which I just don't understand... can anyone see where I've gone wrong here, or are the answers wrong?
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    Maybe you wrongly solved the double fraction - if there was one?
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    (Original post by Kyta)
    Hi guys, I'm doing A2 Physics for OCR, and I've encountered a past paper question about a trolley attached to a spring (which is attached to a vertical unmoving surface) on a smooth table that follows simple harmonic motion, which is as follows;

    'Using your knowledge of Hooke's Law and Newton's second law, determine the period T of the trolley in terms of the force constant k of the spring and the mass m of the trolley.'

    (w = omega)

    Now, I did F = kx = ma; I then put a = w^2x into it, to obtain kx = mw^2x; I cancelled the x on both sides, and then divided by m to obtain w^2 = k/m, as in the answers.

    I then square rooted both sides, flipped both sides over so that T was on top for w, and then multiplied through so that T = 2pi(sqrt(m/k)).

    The answer, however, says that I should have got T = 2pi(sqrt(k/m)), which I just don't understand... can anyone see where I've gone wrong here, or are the answers wrong?
    The answer is wrong, your result is correct. If their answer was correct, \sqrt{k/m} would have the dimension of time. However, \left[ k\right]=MT^{-2}, \left[ m \right] =M, so that \sqrt{\left[ k/m\right]}=T^{-1} (inverse of time). That means that your answer is correct, as it has the dimension of time.
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    Ah, cool, thanks a lot! Glad to have been right :P
 
 
 
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