Question

A 0.40-kg mass is attached to a spring with a force constant of k = 277 N/m, and the mass–spring system is set into oscillation with an amplitude of A = 3.0 cm. Determine the following. (a) mechanical energy of the system J (b) maximum speed of the oscillating mass m/s (c) magnitude of the maximum acceleration of the oscillating mass

Answer 1

To solve this problem it is necessary to apply the concepts related to the kinetic energy expressed in terms of simple harmonic movement, as well as the concepts related to angular velocity and acceleration and linear acceleration and velocity.

By definition we know that the angular velocity of a body can be described as a function of mass and spring constant as

$\omega = \sqrt{\frac{k}{m}}$

Where,

k = Spring constant

m = mass

From the given values the angular velocity would be

$\omega = \sqrt{\frac{277}{0.4}}$

$\omega = 26.31rad/s$

The kinetic energy on its part is expressed as

$E = \frac{1}{2} m\omega^2A^2$

Where,

A = Amplitude

$\omega$ = Angular Velocity

$m = Mass$

PART A) Replacing previously given values the energy in the system would be

$E = \frac{1}{2} m\omega^2A^2$

$E = \frac{1}{2} (0.4)(26.31)^2(3*10^{-2})^2$

$E= 0.1245J$

PART B) Through the amplitude and angular velocity it is possible to know the linear velocity by means of the relation

$v = A\omega$

$v = (3*10^{-2})(26.31)$

$v = 0.7893m/s$

PART C) Finally, the relationship between linear acceleration and angular velocity is subject to

$a = A\omega^2$

$a = (3*10^{-2})(26.31)^2$

$a = 20.76m/s^2$