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3 years ago

11

An object is spun around in circular motion such that its frequency is 500 Hz.

1 answer:

Marrrta [24]3 years ago

7 0

Explanation:

Given parameters:

Frequency = 500Hz

Unknown:

a. Period of rotation

b. Time required to complete 7 rotation

Solution:

The period of rotation is the time taken to complete a turn.

P = $\frac{t}{n}$

P is the period

t is the time

n is the number of turns

Period is the inverse of frequency;

P = $\frac{1}{500}$ = 0.002s

b.

Time required to complete 7 rotation;

P = $\frac{t}{n}$

t = P x n

t is the time

P is the period

n is the number of rotations

t = 0.002 x 7 = 0.014s

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The motion of the mass-spring system is represented by the following formula:

$x(t) = A\cdot \cos (\omega \cdot t + \phi)$

Where:

$x(t)$ - Position of the mass with respect to the equilibrium position, measured in centimeters.

$A$ - Amplitude of the mass-spring system, measured in centimeters.

$\omega$ - Angular frequency, measured in radians per second.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

$a (t) = -\omega^{2}\cdot A \cdot \cos (\omega\cdot t + \phi)$

Where the maximum acceleration of the system is represented by $\omega^{2}\cdot A$ .

The natural frequency of the mass-spring system is:

$\omega = \sqrt{\frac{k}{m} }$

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

If $k = 12\,\frac{N}{m}$ and $m = 0.40\,kg$ , the natural frequency is:

$\omega = \sqrt{\frac{12\,\frac{N}{m} }{0.40\,kg} }$

$\omega \approx 5.477\,\frac{rad}{s}$

Lastly, the maximum acceleration of the system is:

$a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)$

$a_{max} = 359.970\,\frac{cm}{s^{2}}$

The maximum acceleration of the system is 359.970 centimeters per square second.

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