大學物理簡諧運動習題算式 (英)－fksnlix的部落格

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大學物理簡諧運動習題算式 (英)

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一An object undergoing simple harmonic motion takes [0.24s] to travel from one point of zero velocity to the next such point.The distance between those points is [32cm].Calculate the(a)period(b)frequency(c)amplitude of the motion二.What is the maximum acceleration of a platform that oscillates at... 顯示更多一 An object undergoing simple harmonic motion takes [0.24s] to travel from one point of zero velocity to the next such point.The distance between those points is [32cm].Calculate the (a)period (b)frequency (c)amplitude of the motion 二. What is the maximum acceleration of a platform that oscillates at amplitude [2.5cm] and frequency [6.6Hz]? 三. Find the mechanical energy of a block-spring system having a spring constant of [1.8N/cm] and an socillation amplititude of [2.4cm] 四. An oscillating block-spring system has a mechanical energy of [1.00J],An amplitude of [10.0cm],and a maximum speed of [1.20m/s] (a)the spring constant (b)the mass of the block (c)the frequency of oscillation

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一.An object undergoing simple harmonic motion takes t=0.24s to travel from one point of zero velocity to the next such point. The distance between those points is x=32cm. Calculate the(a) period=T=?T=0.24*4=0.96(s) (b) frequencyf=1/T=1.04(Hz) (c) amplitude of the motionA=32*2=64(cm) 二.What is the maximum acceleration of a platform that oscillates at amplitude A=2.5cm and frequency f=6.6Hz?w=2pi*f=2*pi*6.6=41.469(rad/s) a(max)=A*w^2=2.5*41.469^2=4299(cm/s^2)=42.99(m/s^2) 三.Find the mechanical energy of a block-spring system having a spring constant of k=1.8N/cm and an socillation amplititude of A=2.4cmE=k*A^2/2=(1800/100)*(2.4/100)^2=0.010368(J) 四.An oscillating block-spring system has a mechanical energy of E=1.00J, an amplitude of A=0.100m, and a maximum speed of Vmax=1.20m/s(a) the spring constantk=2E/A=2*1/0.1=20(N/m) (b) the mass of the blockVmax=A*√(k/m)m=kA^2/Vm^2=20*0.01/1.44=0.1389(kg) (c)the frequency of oscillationT=2*pi√(m/k)=2*pi√(0.1389/20)=0.5236(s)f=1/T=1.91(Hz)

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