To derive the formulas for the major characteristics of motion as functions of time for a horizontal spring oscillator and to practice using the obtained formulas by answering some basic questions.
A block of mass m is attached to a spring whose spring constant is k. The other end of the spring is fixed so that when the spring is unstretched, the mass is located at x=0. . Assume that the +xdirection is to the right.
The mass is now pulled to the right a distance Abeyond the equilibrium position and released, at time t=0, with zero initial velocity.
Assume that the vertical forces acting on the block balance each other and that the tension of the spring is, in effect, the only force affecting the motion of the block. Therefore, the system will undergo simple harmonic motion. For such a system, the equation of motion is
a(t)=-\frac{k}{m}x(t),
and its solution, which provides the equation for x(t), is
At what time t_1 does the block come back to its original equilibrium position (x=0) for the first time?
Express your answer in terms of some or all of the variables: A, k, and m.

Simple harmonic motion is described as the periodic motion of a point along a straight line with an acceleration that is always toward a fixed point on that line and a distance from that point that is proportional to that acceleration.

The equation of motion for the simple harmonic oscillation of spring mass system is,

a(t) = -k/m x(t)

here, a(t) = is the acceleration at any time

k = is the spring constant, and

m = mass

The equation for displacement is given as,

x(t) = A cos ([tex]\sqrt{k/m}[/tex])t

here,

x(t) = is the displacement at any time.

A = is the amplitude of oscillations,

The velocity is given by,

v(t) = dx(t)/dt

The general expression of velocity for a simple harmonic motion is,

v(t) = v(max) sin ([tex]\sqrt{k/m}[/tex])t

v(max) = is the maximum velocity.

The kinetic energy is given as,

K(t) = 1/2 m[v(t)]²

Now, use the equation of position:

or, x(t) = A cos ([tex]\sqrt{k/m}[/tex])t

or, 0 = A cos ([tex]\sqrt{k/m}[/tex]) t₁

or, t₁ = π/2( [tex]\sqrt{k/m}[/tex])

Thus, the position of particle at time t₁ is 0. The cosine function is zero first time at angle π/2. The time can be calculated by substituting the values in the position function.

To know more about simple harmonic motion refer to:

https://brainly.com/question/20885248

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