By Matthew Hussey (auth.)

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13) Thus the frequency and the period are determined directly by the parameters of the system and are independent of the initial conditions. 3 The displacement of the mass as a function of time. The amplitude A of the oscillation and the period T are shown. Also the initial value of the displacement helps to determine the value of the phase angle -y. 1 has a period of oscillation of 6 s. The mass is 10 kg. Find the frequency and the spring constant. If the mass is doubled what will the period be?

The mass is 10 kg. Find the frequency and the spring constant. If the mass is doubled what will the period be? 4 N/m . 5 s . 4 The velocity x versus displacement x plots for the simple harmonic motion of the mass in the oscillator. The loops are ellipses of axes x (max) or A and f(max) or w 0 A. The radial line 00 rotates clockwise at w0 rad/s. 10 x = A cos (w0 t + 'Y) the velocity may be derived by differentiation, thus x= -woA sin (w 0 t + 'Y) = woA cos ( w0 t + 'Y + ; ) . 14) The velocity is therefore sinusoidal with the same frequency fo as the displacement but is -rr/2 rad ahead of the displacement.

It takes 3 s for the mass to again reach the same phase of the motion. Find the magnitude of the mass, the total mechanical energy in the system and the amplitude of the vibration. Solution The period T and = 3 s. 1 m. 1 J . So the total energy in the system is the sum of these two This total energy is conserved so that when x V2CA 2 • Therefore A 2. 49 m, = A, Ek = 0 and Ep = £ 1 = the amplitude of the vibration. 1, are physically realisable. 6 A simple pendulum, consisting of mass M attached to a string of length L the other end of which is fixed at P.

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