A 3.70-kg object is attached to a spring and placed on frictionless, horizontal

Question

A 3.70-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 =

Explanation / Answer

refer this example: A 1.00 kg object is attached to a spring and placed on a horizontal, smooth surface. A horizontal force of 23.0 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 with an initial position of xi = 0.200 m, and it subsequently undergoes simple harmonic oscillations. (a) Find the force constant of the spring. (b) Find the frequency of the oscillations. (c) Find the maximum speed of the object. Where does this maximum speed occur? (d) Find the maximum acceleration of the object. Where does it occur? (e) Find the total energy of the oscillating system. (f) Find the speed of the object when its position is equal to one third of the maximum value. (g) Find the acceleration of the object when its position is equal to one third of the maximum value. Solution: a) By F = -kx [-ve is just indicating the direction of Force] =>23 = k x (0.2)^2 =>k = 575 N/m (b) By f = 1/2p x v[k/m] =>f = 1/(2 x 3.14) x v[575/1] =>f = 3.82 Hz (c) By PE(spring) = KE(spring) at the equilibrium position i.e. at the +0.20 m from the origion =>1/2mv^2 = 1/2kx^2 =>v^2 = 575 x (0.20)^2 =>v = v23 =>v(max) = 4.80 m/s (d) a(max) = -A?^2 [-ve just indicating the direction]at origin & +0.40 m position =>a(max) = 0.2 x (2 x p x f)^2 =>a(max) = 0.2 x (2 x 3.14 x 3.82)^2 =>a(max) = 115.10 m/s^2 (e) By PE(spring)[max] = 1/2kx^2 =>U = 1/2 x 575 x (0.20)^2 =>U = 11.50 J (f) By U = 1/2mv^2 + 1/2kx^2 =>11.5 = 1/2 x 1 x v^2 + 1/2 x 575 x (0.2/3)^2 =>v = v20.44 =>v = 4.52 m/s (g) By a = -A?^2 =>a = -(0.2/3) x (2 x p x f)^2 =>a = -38.37 m/s^2

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