Question

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.

Answer 1

To find the time t1 when the block comes back to its original equilibrium position (x=0) for the first time, we need to find the period of motion.

Step-by-step explanation:

The equation of motion for a horizontal spring oscillator is given by a(t) = -k/m * x(t), where a(t) is the acceleration, k is the spring constant, m is the mass, and x(t) is the displacement. To find the time t1 when the block comes back to its original equilibrium position (x=0) for the first time, we need to find the period of motion.

The period T of the motion is given by T = 2π√(m/k). Since the block is pulled to the right a distance A beyond the equilibrium position, it will take half a period of time to reach x = 0 again. Therefore, t1 = T/2 = π√(m/k).