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Question

A spring of a 1.00-kg mass with spring constant (k = 25 N/m) is set up horizontally on a frictionless table. The mass is pulled so that the spring is stretched 0.100 m from the equilibrium point as shown in the Figure and released from rest at t = 0. Determine:

(a) the amplitude of the horizontal oscillation A;

(b) the period and frequency from the mass and the spring constant.

(c) the magnitude of the maximum velocity v_max;

(d) the magnitude of the maximum acceleration of the mass.

(e) the total energy.

Answer 1

Let

m = 1kg be the mass attached to the spring,

k = 25N/m be the spring constant,

A = 0.1m be the initial displacement from equilibrium.

(a) The amplitude remains A.

(b) The formula for the period is

T = 2π√(m/k) = 2π√(1kg / 25 kg/s²) = 1.26 s

The frequency is the inverse of period:

f = 1/T = 0.8s^-1

(c) Maximum velocity is achieved as the mass passes the equilibrium position,

when all of the spring energy (k A²/2) is in the kinetic energy (m v_max²/2).

k A²/2 = m v_max²/2

v_max = A√(k/m) = 0.1 m x √(25 kg/s² / 1 kg) = 0.5 m/s

(d) Maximum acceleration occurs at maximum displacement A,

where the force ( k A) is maximized.

By Newton's second law the acceleration

a = k A/m = 25 kg/s² x 0.1 m / 1 kg = 2.5 m/s²

(e) The total energy is the energy of the spring

E = kA²/2 = 25 kg/s² x 0.1² m² / 2 = 0.125 J