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soldi70 [24.7K]
3 years ago
10

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
Physics
1 answer:
andriy [413]3 years ago
6 0

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} }= 0.448s

L_{1}= 10 m

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

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

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}

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Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

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