Shows the position-versus-time graph of a particle in SHM. Positive direction is the direction to the right.
A)At what time or times is the particle moving to the right at maximum speed?
Enter your answers as integers in ascending order separated by commas.
B)At what time or times is the particle moving to the left at maximum speed?
Enter your answers as integers in ascending order separated by commas.
C)At what time or times is the particle instantaneously at rest?
Enter your answers as integers in ascending order separated by commas.

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):

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

where

A is the amplitude

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

So

The velocity of a particle in SHM can be found as the derivative of the displacement, so here we find:

So, the particle is moving to the right at maximum speed when

(because this is the maximum positive value for the cosine part). Solving for t,

We also have another time in which this occurs, when

And also when

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

Which means that this occurs when:

Also when

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:

This occurs when the argument of the cosine is:

Also when

And when

And finally when

So, the particle is at rest at t = 1 s, 3 s, 5 s, 7 s.

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hope this

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west