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Simple Harmonic motion

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圖片參考：http://imgcld.yimg.com/8/n/HA00171827/o/701111080030013873399041.jpg Please answer the questions

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3. Let m be the mass of the block.Loss of potential energy of the block = mg.(0.6 – 0.1) Jwhere m is the acceleration due to gravity, taken to be 10 m/s2 Gain in elastic potential energy of the spring = (1/2).(2400).(0.25-0.1)^2 J = 27 JHence, mg.(0.5) = 27m = 27/0.5g kg = 5.4 kg 4.(a) The two spings in parallel can be replaced by a spring of force constant (2 x 6) kN/m = 12 kN/m. This, together with the spring of 4 kN/m, can be combined to form a spring of force constant (12+4) kN/m = 16 kN/m Thus, when the 35 kg mass is displaced by a displacement x from its equilibrium position, the net force acting on it = 16000x Using, force = mass x acceleration16000x = -35a, where a is the acceleration of the massa = -(16000/35)x = -457xSince acceleration a is proportional to the displacement x and is opposite in direction to that of x, the motion is simple harmonic with angular frequency w = square-root[457] s^-1 (a) Frequency of oscillation = (1/2.pi) x square-root[457] Hz = 3.4 HzPeriod of oscillation = 1/3.4 s = 0.29 s (c ) Since amplitude of oscillation = 20 mm = 0.02 m Maximum acceleration = 457 x 0.02 m/s2 = 9.14 m/s2Maximum velocity = [square-root(457)] x 0.02 m/s = 0.428 m/s

其他解答:8758B59A7FA1EEA7

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