Answer:
A) 0.254 m
B) 1.4563 Hz
C) 0.6867s
Explanation:
Frequency is equal to the angular velocity divided by 2Pi because the units are in radians. Or
=/2
= 9.15/2
=1.4563 Hz
The period is 1/
or 1/1.4563
=0.6867s
(A) The amplitude of this motion is of 0.254 m.
(B) The required frequency of motion is 1.46 Hz.
(C) The time period of motion is 0.68 s.
Given data:
The radius of x-y plane is, r = 0.254 m.
The magnitude of angular velocity, [tex]\omega = 9.15 \;\rm rad/s[/tex].
(A)
The maximum extent of vibration or oscillation that is measured from the equilibrium position is known as Amplitude of motion.
In the given problem, the planar motion has the amplitude (A) which is equal to the magnitude of constant radius in a plane. Then,
A = r
A = 0.254 m
Thus, we can conclude that the amplitude of this motion is of 0.254 m.
(B)
The expression for the frequency of motion is given as,
[tex]\omega = 2\pi f[/tex]
here, f is the frequency of motion. Solving as,
[tex]f = \dfrac{\omega}{2\pi}\\\\\\f = \dfrac{9.15}{2\pi}\\\\f = 1.46 \;\rm Hz[/tex]
Thus, we can conclude that the required frequency of motion is 1.46 Hz.
(C)
The time period of motion is the reciprocal of frequency of motion. The expression for the time period of motion is given as,
[tex]f = \dfrac{1}{T}\\\\T = \dfrac{1}{f}\\\\T = \dfrac{1}{1.46}\\\\T = 0.68 \;\rm s[/tex]
Thus, we can conclude that the time period of motion is 0.68 s.
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