A spring of negligible mass stretches 3.00 cm from its relaxed length when a for

Question

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 8.40 N is applied. A 0.490-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x = 5.00 cm and released from rest at t = 0. (Assume that the direction of the initial displacement is positive.) (a) What is the force constant of the spring? 280 N/m (b) What are the angular frequency (?), the frequency, and the period of the motion? ? = 23.9 rad/s f = 3.8 Hz T = .26 s (c) What is the total energy of the system? .35 J (d) What is the amplitude of the motion? 5 cm (e) What are the maximum velocity and the maximum acceleration of the particle? vmax = 1.2m/s amax = 28.6m/s2 All of the above are right (f) Determine the displacement x of the particle from the equilibrium position at t = 0.500 s. ???cm (g) Determine the velocity and acceleration of the particle when t = 0.500 s. (Indicate the direction with the sign of your answer.) ???m/s ???m/s2

Explanation / Answer

i am doing the last two parts .....force constant of the spring=280 N/m....displacement equation is given by s=Acos(wt-x)...A is the amplitude...w is angular frequency...t is time....x is initial phase....A=5cm..w=23.9rad/s...at t=0...s=5...so 5=5cos(0-x)...so x=0...So s=5cos(23.9t) cm...velocity v=ds/dt=-5*sin(23.9t)*23.9=-119.5 sin(23.9t).cm/s....acceleration a=dv/dt=-119.5cos(23.9t)*23.9=-2856.05cos(23.9t)......(f)displacement at t=5s...is s=5cos(23.9*5)=4.96 cm...(g)velocity=-119.5sin(23.9*5)=-14.24 cm/s.....acceleration=-2856.05cos(23.9*5)=-2835.7 cm/s2....negative sign indicates direction is along negative x axis

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