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stich3 [128]
3 years ago
11

If we double the frequency of a system undergoing simple harmonic motion, which of the following statements about that system ar

e true?
a. The angular frequency is doubled.
b. The amplitude is doubled.
c. The period is doubled.
d. The angular frequency is reduced to one-half of what it was.
e. The period is reduced to one-half of what it was.
Physics
1 answer:
AnnyKZ [126]3 years ago
5 0

Answer:

a. The angular frequency is doubled.

e. The period is reduced to one-half of what it was.

Explanation:

Angular frequency is given as;

ω = 2πf

\frac{\omega _1}{f_1} = \frac{\omega _2}{f_2}

when the frequency is doubled

\frac{\omega _1}{f_1} = \frac{\omega _2}{(2f_1)} \\\\\omega _1 = \frac{\omega _2}{2}\\\\\omega _2 = 2\omega _1

Thus, the angular frequency will be doubled.

Amplitude in simple harmonic motion is the maximum displacement.

Frequency is related to period in simple harmonic motion as given in the equation below;

f = \frac{1}{T} \\\\f_1T_1= f_2T_2\\\\T_2 = \frac{f_1T_1}{f_2}

when the frequency is doubled;

T_2 = \frac{f_1T_1}{2f_1} \\\\T_2 = \frac{T_1}{2}

Thus, the period will be reduced to one-half of what it was.

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A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient
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Answer:

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b) i) The friction force for the box in motion is 147.025 N

ii) The acceleration of box is 4.21625 m/s²

Explanation:

The parameters of the box at rest and the floor are;

The mass of the box = 60 kg

The static coefficient friction of the floor = 0.6

The kinetic coefficient friction of the floor = 0.25

Frictional force = Normal force × Friction coefficient

For an horizontal floor and the box laying on the floor, we have;

The normal force = The weight of the box = Mass of the box × Acceleration due to gravity, g

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The weight of the box  = 60 × 9.81 = 588.6 N

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Therefore;

The minimum force required to start moving the box = The static frictional force = Weight of the box × The static coefficient of friction

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The minimum force required to start moving the box = 352.86 N

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The friction force for the box in motion = 588.1 × 0.25 = 147.025 N

ii) The force, F with the box is in motion, is given as follows;

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