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.

Answer 1

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.