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?
x=Acosωt=Acos2πtT,x=Acos \omega t=Acos \frac{2\pi t}{T},x=Acosωt=AcosT2πt,
v=−2πATsin2πtT, ⟹ v=-\frac{2\pi A}{T}sin\frac{2\pi t}{T}, \impliesv=−T2πAsinT2πt,⟹
A=−vT2πsin2πtT,A=-\frac{vT}{2\pi sin\frac{2\pi t}{T}},A=−2πsinT2πtvT,
A=−−0.964⋅0.616.28⋅sin6.28⋅0.050.61=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,A=−6.28⋅sin0.616.28⋅0.05−0.964⋅0.61=0.19 m,
E=kA22=ω2mA22=2π2mA2T2=2⋅3.142⋅m⋅0.1920.612=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),~E=2kA2=2ω2mA2=T22π2mA2=0.6122⋅3.142⋅m⋅0.192=1.91m (J), where mmm is the mass of a block.
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