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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

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Answer:

a). L= meters, g= [tex]\frac{m}{s^{2} }[/tex]

b). L= 5cm  T=0.448

    L= 10 m  T=6.34

c).  Constant= [tex]\frac{2*\pi }{\sqrt{g} }=\frac{2*\pi }{\sqrt{9.8} }=2.007089923[/tex]

Explanation:

a).

[tex]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[/tex]

b).

[tex]L_{1}= 5 cm[/tex], [tex]5cm *\frac{1m}{100 cm} = 0.05 m[/tex]

[tex]T=2*\pi \sqrt{\frac{L}{g} }[/tex]

[tex]T=2*\pi \sqrt{\frac{0.05}{9.8} }= 0.448[/tex]s

[tex]L_{1}= 10 m[/tex]

[tex]T=2*\pi \sqrt{\frac{L}{g} }[/tex]

[tex]T=2*\pi \sqrt{\frac{10}{9.8} }= 6.43[/tex]s

c).

[tex]g= 9.8 \frac{m}{s^{2} }[/tex]

[tex]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}[/tex]