Question

A toy of mass 0.155 kg is undergoing simple harmonic motion (SHM) on the end of a horizontal spring with force constant 305 N/m . When the object is a distance 1.25×10⁻² m from its equilibrium position, it is observed to have a speed of 0.300 m/s .

Part A) What is the total energy of the object at any point of its motion?

E = ? J

Part B) What is the amplitude of the motion?

A = ? m

Part C) What is the maximum speed attained by the object during its motion?

vmax = ? m/s

Answer 1

(B) angular velocity, ω = √(k/m) = √(305N/m / 0.155kg) .. ..ω = 44.36 rad/s

shm velocity ,v = ω√(r² - x²) .. . x = displacement, r = amplitude (radius)
(v/ω)² = r² - x²
r² = (v/ω)² + x²
r² = (0.300m/s / 44.36rad/s)² + (0.0125m)² = 2.019^-4
Amplitude r = √2.019^-4) .. .. ►r = 1.42^-2 m

(A) shm total energy, E = ½m(ωr)²
Et = ½ x 0.155kg x (44.36 x 1.42^-2)² .. .. ►Et = 3.0^-2 J

(C) v(max) occurs at displacement x = 0 (equilibrium position)
v = ω√(r² - x²) → v(max) = ωr .. .. r = amplitude
v(max) = (44.36rad/s x 1.42^-2 m) .. .. ►v(max) = 0.62 m/s