A 2.7 kg mass is connected to a spring (k=159 N/m) and is sliding on a horizontal frictionless surface. The mass is given an initial displacement of +19 cm and released with an initial velocity of -13 cm/s. Determine the acceleration of the spring at t=3.4 seconds.

The acceleration of the spring at is -29339.947 centimeters per square second.

Explanation:

The mass-spring system experiments a Simple Harmonic Motion, whose kinematic expression is the following:

(1)

Where:

- Position, in centimeters.

- Amplitude, in centimeters.

- Spring constant, in newtons per meter.

- Mass, in kilograms.

- Phase, in radians.

By Differential Calculus, we derive expression for the velocity and acceleration of the mass-spring system:

(2)

(3)

Where and are the velocity and acceleration of the system, in centimeters per second and centimeters per square second, respectively.

If we know , , , and , then we have the following system of nonlinear equations:

(1)

(2)

If we divide (2) by (1):

By (1), we get the value of the amplitude:

If we know that , , , and , then the acceleration of the spring is:

The acceleration of the spring at is -29339.947 centimeters per square second.

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explanation: