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jek_recluse [69]

1 year ago

9

The position of an object that is oscillating on a spring is given by the equation x = (17.4 cm) cos[(5.46 s-1)t]. what is the a

ngular frequency for this motion?

1 answer:

bulgar [2K]1 year ago

6 0

The angular frequency of this motion is 5.46 rad/s.

The oscillation of spring is an example of Simple Harmonic Motion(SHM).

The general equation of an SHM is given by the formula.

X = Acos(wt)

Here A is the amplitude

ω is the angular frequency

T is the time

Comparing the above equation with the given condition,

X = 17.4 cm cos(5.46t)

A = 17.4 cm

ω = 5.46 rad/s

T = 1 s

Hence, the angular frequency of this motion is 5.46 rad/s.

To know more about the "general equation of SHM", refer to the link below:

brainly.com/question/14869852?referrer=searchResults

#SPJ4

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Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

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