A 1.10-kg object is attached to a spring and placed on frictionless, horizontal

Question

A 1.10-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 5.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis). The object is now released from rest from this stretched position, and it subsequently undergoes simple harmonic oscillations. (a) Find the force constant of the spring. N/m (b) Find the frequency of the oscillations. Hz (c) Find the maximum speed of the object. m/s (d) Where does this maximum speed occur? x =

Explanation / Answer

A horizontal force of 11.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position a) this means spring constant (k) can be calculated using kx=F k*0.2 = 11 or k = 55 b) we know frequency f= 1/2pi sqrt[(k/m)] = 1/6.28 * sqrt(55/2.2) = 0.796 Hz c) maximum speed = maximum kinetic energy = which occurs at equilibrium position = maximum spring potential energy 1/2*m*v^2 = 1/2* k * x^2 2.2 * v^2 = 55* (0.2)^2 so maximum speed, v = 1 m/s d) like said earlier maximum speed occurs at equilibrium position i.e., at x=0 m e) maximum acceleration occurs at ends. maximum acceleration = max force / mass = 11/2.2 = 5m/s^2 f) max acceleration occurs at x= -0.2m g) total energy of oscillating system is constant = Kinetic energy at equilibrium = spring potential energy at extreme. = 1/2 k x^2 = 1/2 * 55 * (0.2)^2 = 1.1 let me know if you have any doubts

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