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

Answer:

a). L= meters, g= $\frac{m}{s^{2} }$

b). L= 5cm  T=0.448

    L= 10 m  T=6.34

c).  Constant= $\frac{2*\pi }{\sqrt{g} }=\frac{2*\pi }{\sqrt{9.8} }=2.007089923$

Explanation:

a).

$T= 2*\pi \sqrt{\frac{L}{g} } = 2*\pi \sqrt{\frac{m}{\frac{m}{s^{2} } } } \\T= 2*\pi \sqrt{\frac{s^{2}*m }{m} }=2*\pi \sqrt{s^{2} } \\T= s$

b).

$L_{1}= 5 cm$, $5cm *\frac{1m}{100 cm} = 0.05 m$

$T=2*\pi \sqrt{\frac{L}{g} }$

$T=2*\pi \sqrt{\frac{0.05}{9.8} }= 0.448$s

$L_{1}= 10 m$

$T=2*\pi \sqrt{\frac{L}{g} }$

$T=2*\pi \sqrt{\frac{10}{9.8} }= 6.43$s

c).

$g= 9.8 \frac{m}{s^{2} }$

$T=2*\pi *\frac{\sqrt{L} }{\sqrt{g} } =T=2*\pi *\frac{\sqrt{L} }{\sqrt{9.8} } \\T= 2*\pi \frac{1}{\sqrt{9.8}} *\sqrt{L}\\T= 2.007089923*\sqrt{L}$