Will give brainliest!

1. you have a simple harmonic oscillator. at what position(s) during simple harmonic motion is an oscillator's speed zero?

check all that apply.
a) at the equilibrium position
b) at t = 0
c) at the maximum distances from equilibrium
d) because the oscillator is constantly moving, there is no place where its speed is zero.

2. you have a simple harmonic oscillator. where is its acceleration zero?

check all that apply.
a) at t = 0
b) at the equilibrium position
c) at the maximum distances from equilibrium
d) because the oscillator is constantly moving, there is no place where its acceleration is zero.

3. you have a simple harmonic oscillator. where is the magnitude of its acceleration at its maximum?

check all that apply.
a) at the maximum distances from equilibrium
b) at t = 0
c) at the equilibrium position
d) the acceleration is constant.

4. why an object attached to a light spring undergoes simple harmonic motion after it is displaced from the equilibrium position?

a) because the position-versus-time graph is a sinusoidal-type function.
b) because the acceleration of the object is proportional to its displacement with a negative sign.
c) because the motion is periodic and has a constant period.
d) a and b are correct.

(1) At what position(s) during simple harmonic motion is an oscillator's speed zero?

(C) at the maximum distance from equilibrium

Explanation:

The speed (and the kinetic energy) of the oscillating mass is zero when it reaches the farthest point away from its center point (equilibrium). This can be shown by deriving the speed formula and analyzing it:

Let s(t) be the function of the position that follows a simple sine oscillation with an angular frequency omega:

the derivative of this function is the instantaneous rate of change of distance in time, i.e., the velocity:

Suppose the frequency is 1Hz. As you can see, the cosine (and so the speed) is minimum, namely zero, at pi/2, 3pi/2, 5pi/2, etc. Those are points that align with the s(t) function being at the extreme points of its amplitude.

(2) Where is its acceleration zero?

B) At the equilibrium position

Explanation:

Let us derive the acceleration from the velocity function above:

As you can see the acceleration also follows a sine wave but in exactly opposite direction to the location. For example, when the mass is up at +A, the acceleration points down with amplitude  , and vice versa (we will use this also to answer question (4).

The above function is zero at t=0, pi, 2pi, etc. (again assuming frequency is 1Hz in this example). Those are points that correspond to the mass being in its equilibrium location, as per s(t).

(3)  Where is the magnitude of its acceleration at its maximum?

A) At the maximum distances from equilibrium

Explanation:

The function a(t) (see 2 above) has maxima at t=pi/2, 3pi/2, etc. (with frequency 1Hz), which correspond to locations at the extreme amplitudes, as seen on the s(t) function at the same points.

(4)  Why an object attached to a light spring undergoes simple harmonic motion after it is displaced from the equilibrium position?

B) Because the acceleration of the object is proportional to its displacement with a negative sign.

Explanation:

This is the key relationship. Initially, the system must be given energy to move out of its equilibrium and have a non-zero displacement. This will induce a counteracting acceleration that will make the mass move through its equilibrium in an oscillating manner.

Please note that (A) is probably a tempting but incorrect answer, as that is not a cause but rather an effect.

a.   mass number from atomic number