Question #167429

A block attached to a spring is released at t =0 with the spring extended.the period of oscillation is 0.61s.At t = 0.05s the velocity is v=-96.4cm/a.(1)what is the amplitude? (2)what is the total energy?


1
Expert's answer
2021-02-28T07:33:17-0500

x=Acosωt=Acos2πtT,x=Acos \omega t=Acos \frac{2\pi t}{T},

v=2πATsin2πtT,    v=-\frac{2\pi A}{T}sin\frac{2\pi t}{T}, \implies

A=vT2πsin2πtT,A=-\frac{vT}{2\pi sin\frac{2\pi t}{T}},

A=0.9640.616.28sin6.280.050.61=0.19 m,A=-\frac{-0.964\cdot 0.61}{6.28\cdot sin\frac{6.28\cdot 0.05}{0.61}}=0.19~m,

E=kA22=ω2mA22=2π2mA2T2=23.142m0.1920.612=1.91m (J), E=\frac{kA^2}{2}=\frac{\omega ^2 m A^2}{2}=\frac{2\pi ^2mA^2}{T^2}=\frac{2\cdot 3.14^2\cdot m\cdot 0.19^2}{0.61^2}=1.91m~(J),~ where mm is the mass of a block.


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