A particle with mass 2.51 kg oscillates horizontally at the end of a horizontal

Question

A particle with mass 2.51 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.953 m and a duration of 125 s for 77 cycles of oscillation. Find the frequency, f, the speed at the equilibrium position, vmax, the spring constant, k, the potential energy at an endpoint, Umax, the potential energy when the particle is located 47.1% of the amplitude away from the equiliibrium position, U, and the kinetic energy, K, and the speed, v, at the same position. Please help!!

Explanation / Answer

T=125/77=1.6234 s=2*pi/w ...w=3.87 rad/s =sqrt(k/m) ....k=38.6 N/m ....f=1/T=0.616 Hz.....

the speed at the equilibrium position= vmax=Aw =0.953*3.87=3.688 m/s

Umax at the end point=0.5*k*A^2=0.5*38.6*0.953^2=17.53 J

x=0.4488=47.1% of the amplitude away from the equiliibrium position.....U=0.5*k*x^2=0.5*38.6*0.4488^2=3.887 J

Umax-U =K=17.53-3.887=13.643 J= 0.5*m*v^2....v=3.287 m/s

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