Question

The period T of a simple pendulum depends on the length L of the pendulum and the acceleration of gravity g (dimensions L/P). (a) Find a simple combination of L and g that has the dimensions of time. (b) Check the dependence of the period T on the length L by measuring the period (time for a complete swing back and forth) of a pendulum for two different values of L. (c) The correct formula relating T to L and g involves a constant that is a multiple of 1 T, and cannot be obtained by the dimensional analysis of Part (a). It can be found by experiment as in Part (b) if g is known. Using the value g

Answer 1

a). L= meters, g=
`(m)/(s^(2) )`

b). L= 5cm T=0.448

L= 10 m T=6.34

c). Constant=
`(2*\pi )/(√(g) )=(2*\pi )/(√(9.8) )=2.007089923`

Step-by-step explanation:

a).

`T= 2*\pi  \sqrt{(L)/(g) } = 2*\pi \sqrt{(m)/((m)/(s^(2) ) ) } \\T= 2*\pi \sqrt{(s^(2)*m )/(m) }=2*\pi  \sqrt{s^(2)  } \\T= s`

b).

`L_(1)= 5 cm`,
`5cm *(1m)/(100 cm) = 0.05 m`

`T=2*\pi \sqrt{(L)/(g) }`

`T=2*\pi \sqrt{(0.05)/(9.8) }= 0.448`s

`L_(1)= 10 m`

`T=2*\pi \sqrt{(L)/(g) }`

`T=2*\pi \sqrt{(10)/(9.8) }= 6.43`s

c).

`g= 9.8 (m)/(s^(2) )`

`T=2*\pi *(√(L) )/(√(g) ) =T=2*\pi *(√(L) )/(√(9.8) ) \\T= 2*\pi (1)/(√(9.8)) *√(L)\\T= 2.007089923*√(L)`