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A pendulum of length 77.5 cm swings back and forth at a location on earth where g = 9.82 m/s². The frequency of this oscillation, in Hz, is most nearly_____.


Sagot :

Answer:

0.57 Hz

Explanation:

Frequency

In physics, frequency measures how many oscillations occur in a unit of time, usually in hertz (Hz), cycle per second.

Frequency is also the inverse of period, which measures the duration to complete an oscillation.

                                                 [tex]T=\dfrac{1}{f}[/tex]

, where T is period.

[tex]\hrulefill[/tex]

Solving the Problem

We can calculate the pendulum's period and use the relationship between period and frequency to find our final answer!

To find the period of the pendulum a formula can be used

                                          [tex]T=2\pi\sqrt{\dfrac{L}{g}[/tex],

where L is the length of the string (in meters) and g is the gravitational acceleration of the environment (in meters per second squared).

Plugging in the given value L and g,

                             [tex]T=2\pi\sqrt{\dfrac{0.775}{9.82} }=1.765123007[/tex].

So,

                                [tex]f=\dfrac{1}{T}=0.5665=0.57\:Hz[/tex].