Answer to Question #325726 in Physics for Claire

Question #325726

A 0.25-kg mass at the end of a spring oscillates 2.2 times per second with an amplitude of 0.15 m. Determine (a) the speed when it passes the equilibrium point, (b) the speed when it is 0.10 m from equilibrium, (c) the total energy of the system, and (d) the equation describing the motion of the mass, assuming that at t=0, x was a maximum.


1
Expert's answer
2022-04-11T12:05:57-0400

The angular frequency of oscillation

ω=2πf=23.142.2=13.8rad/s\omega=2\pi f\\ =2*3.14*2.2=13.8\:\rm rad/s

(a)

vmax=ωA=13.80.15=2.1  m/sv_{\max}=\omega A\\ =13.8*0.15=2.1\;\rm m/s

(b)

v=vmax2ω2x2=2.1213.820.102=1.6m/sv=\sqrt{v_{\max}^2-\omega^2x^2}\\ =\sqrt{2.1^2-13.8^2*0.10^2}=1.6\:\rm m/s

(c)

E=mω2A22=0.2513.820.1522=0.54JE=\frac{m\omega^2A^2}{2}\\ =\frac{0.25*13.8^2*0.15^2}{2}=0.54\:\rm J

(d)

x=0.15cos(13.8t)x=0.15\cos(13.8t)


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