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A 200-g mass is attached to a spring of constant k=5.6N/m and set into oscillation with amplitude A=25cm. Determine (a)the frequency in hertz, (b)the period, (c)the maximum velocity and (d)the maximum force in the spring.(a)0.842Hz(b)1.19s(c)1/32 m/s(d)1.4NA particle undergoes simple harmonic motion with... 顯示更多 A 200-g mass is attached to a spring of constant k=5.6N/m and set into oscillation with amplitude A=25cm. Determine (a)the frequency in hertz, (b)the period, (c)the maximum velocity and (d)the maximum force in the spring. (a)0.842Hz(b)1.19s(c)1/32 m/s(d)1.4N A particle undergoes simple harmonic motion with maximum speed 1.4m/s and maximum acceleration 3.1 m/s^2. Find (a)the angular frequency, (b)the period, and (c)the amplitude of the motion. (a)2.21s^(-1) (b)2.84s(c)0.632m A 450-g mass on a spring is oscillation at 1.2Hz. The total energy of the oscillation is 0.51J. What is the amplitude? 20cm 更新: 我第二題的答案那是什麼意思阿? 2.21s^(-1) 角頻率也能表示成這樣嗎?第一次看到這種寫法@@

最佳解答:

1. X = a*sin(w*t) , V = a*w*cos(w*t) where a = 0.25 m , w = (5.6/0.2)^(1/2) = 5.292 rad/s (a) f = w/(2*pi) = 5.292/(2*3.14) = 0.842 Hz (b) T = 1/f = 1.19 s (c) Vmax = a*w = 0.25*5.292 = 1.32 m/s (d) Fmax = k*Xmax = 5.6*0.25 = 1.4 N 2. X = a*sin(w*t) => V = a*w*cos(w*t) , A = -a*w^2*sin(w*t) Vmax = a*w = 1.4 , Amax = a*w^2 = 3.1 (a) w = Amax/Vmax = 3.1/1.4 = 2.21 rad/s (b) T = 2*pi/w = 2*3.14/2.21 = 2.84 s (c) a = 1.4/w = 1.4/2.21 = 0.633 m 3. X = a*sin(2*pi*f*t) => V = a*2*pi*f*cos(2*pi*f*t) , f = 1.2 Hz (1/2)*m*Vmax^2 = 0.51 => (1/2)*0.45*(a*2*pi*1.2)^2 = 0.51 => 12.778*a^2 = 0.51 => a = 0.20 m = 20 cm

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