Help!!!!!!!!!!?

Question

Do you think you can help me on my physics homework? I got stuck on number two and the problem reads: The period of a simple pendulum, defined as the time for one complete oscillation, is measured in time units and given by T equals two pi square root of l over g where l is the lenghth of the pendulum and g is the acceleration due to gravity, in units of lenght divided by time squared. Show that this equation is dimensionally consistent.

Answer 1

We just need to put the SI units in for each variable and see that it ends up correct after simplification.

T = 2π√(l/g)

- Units of T are seconds (s)
- Units of l are meters (m)
- Units of g are meters per second squared (m/s²)
- 2π is just a constant, so it is unitless

So, let's just plug in these units and simplify:

s = √(m/(m/s²))
==> meters cancel out
s = √(1/(1/s²))
==> make positive exponent negative by putting it in the numerator
s = √(1/(s^(-2)))
==> now, again put exponent in numerator by changing sign of exponent
s = √(s²)
==> exponent and square root cancel
s = s

We have reduced it to a true statement, so you have shown that it is dimensionally consistent.

Answer 2

As you have said
T=2 pi sqrt(l/g)

T- units of time (sec, min, hours) we use seconds)
l - length (meters, feet, miles, mm..) lets use meters
g - acceleration - length / (time)^2

now substitute

T= sec = 2 pi sqrt( length/(length/ sec^2))
T=sec= 2pi sqrt(sec)^2)
sec= 2 pi sec

pi is dimensionless constant

so sec=sec

and T=2 pi sqrt(l/g) is dimensionally consistent