The period of a simple pendulum, defined as the time necessary for one complete oscillation, is measured in time units and is given by T=2πl

Question

The period of a simple pendulum, defined as the time necessary for one complete oscillation, is measured in time units and is given by T=2πlg where l is the length of the pendulum and g is the acceleration due to gravity, in units of length divided by time squared. Show that this equation is dimensionally consistent.

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Khánh Gia 2 months 2021-07-29T20:36:39+00:00 1 Answers 10 views 0

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    2021-07-29T20:38:32+00:00

    Answer:

    Explanation:

    T = 2π √l/g

    The dimension for l = m

    The dimension for g = m/s²

    The dimension for 2π is nothing. Since it’s a constant, it is dimensionless.

    Now we proceed ahead. Since we are not using the 2π, for the sake of this proving, our formula will temporarily be written as

    T = √l/g

    Inputting the dimensions, we have

    T = √(m) / (m/s²)

    T = √(m * s²/m)

    T = √s²

    T = s

    Since the unit of period itself is in s, we can adjudge that the equation is dimensionally constant.

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