Respuesta :
a) The displacement at t = 3.0s is x = 0
b) The velocity at t = 3.0s is [tex]-15\pi[/tex] m/s
c) The acceleration at t = 3.0s is a = 0
d) The phase of motion is [tex]\phi = \frac{\pi}{2}[/tex] [rad]
e) The frequency is 2.5 Hz
f) The period of the motion is 0.4 s
Explanation:
a)
The displacement of the body in a simple harmonic motion (SHM) is given by an equation in the form
[tex]x(t) = A cos (\omega t + \phi)[/tex]
where
A is the amplitude of the motion
[tex]\omega[/tex] is the angular frequency
t is the time
[tex]\phi[/tex] is the phase
For the body in SHM in this motion, the displacement is given by
[tex]x(t) = (3.0) cos(5\pi+ \frac{\pi}{2}) [m][/tex]
Here we want to find the displacement when
t = 3.0 s
Substituting,
[tex]x(3.0) = (3.0) cos(5\pi (3.0) + \frac{\pi}{2})=0[/tex]
b)
In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:
[tex]v(t) = x'(t)=(A cos(\omega t+\phi))' =-\omega A sin(\omega t+\phi)[/tex]
And so, for the motion in this problem,
[tex]v(t)=-5\pi (3.0) sin(5\pi t+\frac{\pi}{2})[/tex]
Since [tex]\omega=5\pi[/tex] and [tex]\phi = \frac{\pi}{2}[/tex].
Now, substituting t = 3.0 s, we find the velocity of the body at that time:
[tex]v(3.0)=-5\pi (3.0) sin(5\pi (3.0)+\frac{\pi}{2})=-15 \pi m/s[/tex]
c)
In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:
[tex]a(t) = v'(t)=(-\omega A sin(\omega t+\phi))' =-\omega^2 A cos(\omega t+\phi) = -\omega^2 x(t)[/tex]
And so, for the motion in this problem,
[tex]a(t)=-(5\pi)^2 x(t)[/tex]
Since [tex]\omega=5\pi[/tex].
Now, substituting t = 3.0 s, we find the acceleration of the body at that time:
[tex]a(3.0)=-(5\pi)^2 x(3.0) = 0[/tex]
d)
The phase of motion is the angular displacement of the motion at t = 0. In this problem, its value can be inferred by comparing the equation
[tex]x(t) = A cos (\omega t + \phi)[/tex]
with the equation for this motion:
[tex]x(t) = (3.0) cos(5\pi+ \frac{\pi}{2}) [m][/tex]
We immediately see that the phase of motion, [tex]\phi[/tex], is the second term in the argument of the cosine: therefore, it is
[tex]\phi = \frac{\pi}{2}[/tex] [rad]
e)
The frequency of the motion can be inferred by the angular frequency. In fact, we know already that in this motion, the angular frequency is
[tex]\omega = 5 \pi[/tex] rad/s
The angular frequency can be also rewritten as
[tex]\omega = 2\pi f[/tex]
where f is the frequency of the motion.
Therefore, re-arranging the equation for f and substituting the value of [tex]\omega[/tex] that we have, we can find the frequency of the motion:
[tex]f=\frac{\omega}{2\pi}=\frac{5\pi}{2\pi}=\frac{5}{2}=2.5 Hz[/tex]
f)
The period of a simple harmonic motion is the time required for one complete oscillation to occur. It is related to the frequency by the equation
[tex]f=\frac{1}{T}[/tex]
where
f is the frequency
T is the period
For the motion in this problem, the frequency is
f = 2.5 Hz
Therefore, we can re-arrange the equation to find the period:
[tex]T=\frac{1}{f}=\frac{1}{2.5}=0.4 s[/tex]
Learn more about frequency and period:
brainly.com/question/5354733
brainly.com/question/9077368
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