A 1.00-kg object is attached to a spring and placed on frictionless, horizontal

Question

A 1.00-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 13.0 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 = Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m (e) Find the maximum acceleration of the object. m/s^2 (f) Where does the maximum acceleration occur? x = m (g) Find the total energy of the oscillating system. J (h) Find the speed of the object when its position is equal to one-third of the maximum value. Your response is within 10percent of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m/s (i) Find the acceleration of the object when its position is equal to one-third of the maximum value. Your response differs from the correct answer by more than 10percent. Double check your calculations. m/s^2

Explanation / Answer

A 3.60-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 15.0 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. The force constant is F/?x = 15.0/0.200; k = 75.0 N/m (b) f = (1/2p)v[k/m] = (1/2p)v[75/3.6] = 0.726 Hz (c) The displacement is given by A*sin(2pft) A = initial deflection = 0.200. The velocity is the derivative of this or 2pfA*cos(2pft). At max (where cos = 1) the velocity is 2pfA or 0.913 m/s. (d) The acceleration is the derivative of the velocity or -(2pf)²A*sin(2pft) and the value of the max is (2pf)²A = 4.16 m/s² (e) The max acceleration occurs when the velocity is zero, which is when the mass passes through the neutral point of the spring (the position of the string when it is unforced). (f) The total energy is the same as the initial energy, which is 0.5*k*?x² = 1.5 J (g) The position is A*sin(2pft). The time when the position is A/3 is given by 1/3 = sin(2pft) so 2pft = arcsin(1/3). The velocity is 2pfA*cos(2pft) so at A/3 v = 2pfA*cos[arcsin(1/3)] = 2pfA*v8/3 = 0.860 m/s (h) Acceleration is -(2pf)²A*sin(2pft) and at A/3 2pft = arcsin(1/3) sin[arcsin(1/3)] = 1/3 so the acceleration is -(2pf)²A/3 = -164/3 = 1.39 m/s²

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