Question #201914

A 2.00 [kg] frictionless block is attached to an ideal spring of force constant 300. [N/m]. At t = 0 the

block has velocity -4.00 [m/s] and displacement 0.200 [m]. Determine the following:

A.The amplitude of the oscillation


B. The position-time and velocity-time functions of the spring-mass system


1
Expert's answer
2021-06-02T09:36:52-0400

For a spring-mass system without any friction we have the following equation that relates the total energy of the system with the position and velocity of the oscillating mass:


12kx2+12mv2=12kA2    A=mv(t)2k+x(t)2\frac{1}{2}kx^2+\frac{1}{2}mv^2=\frac{1}{2}kA^2\implies A = \sqrt{\frac{mv_{(t)}^2}{k} +x_{(t)}^{2}}


Then, we substitute the velocity and position at the start with the spring constant and the mass to find the amplitude A:


A=(2kg)(4m/s)2200N/m+(0.2m)2=0.4472mA = \sqrt{\frac{(2\,kg)(-4\,m/s)^2}{200\,N/m} +(0.2\,m)^{2}}=0.4472 \,m

B) Then, the position-time and velocity-time functions of the spring-mass system would be


x(t)=Acos(ωt+ϕ)ddtx(t)=v(t)=Aωsin(ωt+ϕ)x(t) = A \cos(\omega t + \phi)\\ \frac{d}{dt}x(t)=v(t) = -A\omega \sin(\omega t + \phi)

with the constants:


  1. A = 0.4472 m
  2. ω=km=300N/m2kg=150s1=12.247s1ω=\sqrt{\frac{k}{m}}=\sqrt{\frac{300\,N/m}{2\,kg}}=\sqrt{150}\,s^{-1} =12.247\,s^{-1}
  3. ϕ=cos1(x(0)A)=63.434°\phi=cos^{-1}(\frac{x_{(0)}}{A})=63.434°


Reference:

- Young, H. D., Freedman, R. A., & Ford, A. L. (2006). Sears and Zemansky's university physics (Vol. 1). Pearson education.


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