A) t = 0 s, 4 s, 8 s
B) t = 2 s, 6 s
C) t = 1 s, 3 s, 5 s, 7 s
Explanation:
A)
The figure is missing: find it in attachment.
A particle is said to be in Simple Harmonic Motion (SHM) when it is acted upon a restoring force proportional to its displacement, and therefore, the acceleration of the particle is directly proportional to its displacement (but in the opposite direction):
[tex]a\propto -x[/tex]
Also, the displacement of a particle in SHM can be described by a sinusoidal function, as shown in the figure for the particle in this problem.
For the particle in this problem, from the figure we can see that the displacement is described by a function in the form
[tex]x(t)=A sin(\omega t)[/tex]
where
A is the amplitude
[tex]\omega=\frac{2\pi}{T}[/tex] is the angular frequency, where T is the period. From the graph, we see that the particle completes 1 oscillation in 4 seconds, so
T = 4 s
So the angular frequency is
[tex]\omega = \frac{2\pi}{4}=\frac{\pi}{2}rad/s[/tex]
So
[tex]x(t)=Asin(\frac{\pi}{2}t)[/tex]
The velocity of a particle in SHM can be found as the derivative of the displacement, so here we find:
[tex]v(t)=x'(t)=A\omega cos(\omega t) = \frac{A\pi}{2}cos(\frac{\pi}{2}t)[/tex]
So, the particle is moving to the right at maximum speed when
[tex]cos(\frac{\pi}{2}t)=+1[/tex]
(because this is the maximum positive value for the cosine part). Solving for t,
[tex]\frac{\pi}{2}t=0\\t=0 s[/tex]
We also have another time in which this occurs, when
[tex]\frac{\pi}{2}t=2\pi\\\rightarrow t=\frac{4\pi}{\pi}=4 s[/tex]
And also when
[tex]\frac{\pi}{2}t=4\pi\\\rightarrow t=8s[/tex]
B)
In this case, we want to find the time t at which the particle is moving to the left at maximum speed.
This occurs when the cosine part has the maximum negative value, so when
[tex]cos(\frac{\pi}{2}t)=-1[/tex]
Which means that this occurs when:
[tex]\frac{\pi}{2}t=\pi\\\rightarrow t=2 s[/tex]
Also when
[tex]\frac{\pi}{2}t=3\pi\\\rightarrow t=6 s[/tex]
So, the particle has maximum speed moving to the left at 2 s and 6 s.
C)
The particle is instantaneously at rest when the speed is zero, this means when the cosine part is equal to zero:
[tex]cos(\frac{\pi}{2}t)=0[/tex]
This occurs when the argument of the cosine is:
[tex]\frac{\pi}{2}t=\frac{\pi}{2}\\\rightarrow t = 1 s[/tex]
Also when
[tex]\frac{\pi}{2}t=\frac{3\pi}{2}\\\rightarrow t = 3 s[/tex]
And when
[tex]\frac{\pi}{2}t=\frac{5\pi}{2}\\\rightarrow t = 5 s[/tex]
And finally when
[tex]\frac{\pi}{2}t=\frac{7\pi}{2}\\\rightarrow t = 7s[/tex]
So, the particle is at rest at t = 1 s, 3 s, 5 s, 7 s.
