In a horizontal block-spring system,the block mass is m=0.5kg and the spring constant is

k =2N/m. At t1=0.1s the block velocity is +0.263m/s and the acceleration is -0.216m/s2. Assuming that the system is undergoing SHM.Write the expression for block displacement from equilibrium point of spring with time

I know we have to find the 3 parameter A, w(angular frequency) and phase

equation : y(t)=Asin(wt+phase)
1.Find angular frequency
w=sqrt(k/m) which is equal to 2rad/s
I really don't know how to find the rest please help me

x= A sin (wt+P

v = Aw cos (wt+P)
a = -Aw^2 sin (wt+P) = -w^2 x

f = m a
-kx = -m w^2 x
so as you probably know
w^2 = k/m

here w^2 = 2/.5 = 4 so w = 2
now
at t = .1
.263 = A(2)cos(.2+P)
-.216 = -A (4)sin (.2+P)

A sin = .054

A cos = .132

sin/cos = tan(.2+P) = .054/.132

inverse tan for .2+P in radians

Well, finding the rest of the parameters involves using the given information about the block's velocity and acceleration at a specific time.

Let's start with finding the amplitude (A) of the block's displacement. Amplitude represents the maximum distance the block moves away from the equilibrium position. We can determine this by looking at the block's velocity at t1=0.1s.

Given that the block's velocity is +0.263m/s, we know that at its maximum displacement, the velocity is zero. This means that it reaches the maximum amplitude when the velocity is changing direction (from positive to negative or vice versa).

Since we have a symmetrical system, we can assume that the displacement is at its maximum when the velocity is zero. Therefore, the amplitude of the block's displacement is equal to the magnitude of its initial velocity: A = 0.263m.

Now, let's move on to find the phase. The phase (Φ) represents the starting point of the motion in the oscillation cycle. To determine this, we need to consider the block's acceleration at t1=0.1s.

Given that the acceleration is -0.216m/s^2, we know that it is changing direction when the block reaches its maximum displacement. Therefore, the block's maximum displacement occurs when the acceleration is zero.

Since we have a symmetrical system, we can assume that the displacement is at its maximum when the acceleration is zero. Therefore, the phase is 180 degrees or π radians, as at maximum displacement, the block's displacement function will be passing through the origin. So Φ = π.

Putting all the values together, the expression for the block's displacement from the equilibrium point of the spring with time will be:

y(t) = 0.263sin(2t + π)

Remember to plug in the appropriate units when using the equation. Good luck with your calculations!

To find the remaining parameters, you need to use the given information about the block's velocity and acceleration at t1=0.1s.

2. Find the amplitude (A):
The amplitude of the block's displacement can be found using the equation:

A = |(V_max|) / w

Where V_max is the maximum velocity of the block. In this case, V_max is the absolute value of the block's velocity, which is 0.263 m/s. Therefore,

A = |0.263| / 2 = 0.1315 m

3. Find the phase angle (Φ):
The phase angle can be determined using the equation:

Φ = arctan(-(a_max) / (wV_max))

Where a_max is the maximum acceleration of the block. In this case, a_max is the negative value of the block's acceleration, which is -0.216 m/s^2. Using the angular frequency (w = 2 rad/s) and V_max (0.263 m/s), we can calculate the phase angle:

Φ = arctan(-(-0.216) / (2*0.263)) = arctan(0.411) ≈ 0.392 radians

Therefore, the expression for the block's displacement from the equilibrium point of the spring with time (t) is:

y(t) = 0.1315 * sin(2t + 0.392)

To find the remaining two parameters (A and phase) in the expression for block displacement, you'll need to use the initial conditions given in the problem:

At t1 = 0.1s, the block has a velocity of +0.263m/s and an acceleration of -0.216m/s^2.

Using these initial conditions, we can determine the values of A (amplitude) and phase.

1. Find the amplitude (A):
The amplitude is the maximum displacement from the equilibrium position. It can be determined using the initial velocity.
v(t) = Awcos(wt + phase) [Using the derivative of the displacement equation]
At t = t1 = 0.1s, v(t) = +0.263m/s
0.263 = Awcos(w * 0.1 + phase)

2. Find the phase angle (phase):
Using the initial acceleration, we can find the phase angle.
a(t) = -A * w^2 * sin(wt + phase) [Using the second derivative of the displacement equation]
At t = t1 = 0.1s, a(t) = -0.216m/s^2
-0.216 = -A * w^2 * sin(w * 0.1 + phase)

Now, you have two equations with two unknowns (A and phase). You can solve these equations simultaneously to find the values of A and phase.