Question 19.QE.3: Equation Jeopardy The position of a vibrating cart attached ......
Equation Jeopardy
The position of a vibrating cart attached to a spring is described by the equation
x=(0.050 \mathrm{~m}) \cos \left(\left(12 \mathrm{~s}^{-1}\right) t\right)
What can you determine about the motion?
Represent mathematically Compare the above equation with the position versus-time equation for simple harmonic motion: x=A \cos \left(\frac{2 \pi}{T} t\right) We see that the amplitude A of the vibration is 0.050 m and \frac{2 \pi}{T}=12 \mathrm{~s}^{-1} .
The period is then T =(2π/12)s = (π/6) s. The frequency of vibration f=1 / T=(6 / \pi) \mathrm{s}^{-1}= (6 / \pi) \mathrm{Hz} The object has its maximum positive displacement at time zero.
Solve and evaluate The velocity of the vibrating cart is
v_x=-\frac{2 \pi}{(\pi / 6) \mathrm{s}}(0.050 \mathrm{~m}) \sin \left(\frac{2 \pi}{(\pi / 6) \mathrm{s}} t\right)
=-(0.60 \mathrm{~m} / \mathrm{s}) \sin \left(\left(12 \mathrm{~s}^{-1}\right) t\right)
The amplitude of the velocity (the cart’s maximumspeed) is 0.60 m/s. The cart’s acceleration is
a_x=-\left(\frac{2 \pi}{(\pi / 6) \mathrm{s}}\right)^2(0.050 \mathrm{~m}) \cos \left(\frac{2 \pi}{(\pi / 6) \mathrm{s}} t\right)
=-\left(7.2 \mathrm{~m} / \mathrm{s}^2\right) \cos \left(\left(12 \mathrm{~s}^{-1}\right) t\right)
The amplitude of the acceleration (the cart’s maximum acceleration) is 7.2 /s².
Try it yourself: Suppose the amplitude of vibration in this example remained the same, but the period was reduced by half. By what factors would this affect the cart’s maximum speed and maximum acceleration? Try to answer without plugging numbers into equations.
Answer: Since the maximum speed is proportional to 1/T, the maximum speed would double if the period were halved. Since the maximum acceleration is proportional to 1/T², themaximum acceleration would quadruple if the period were halved.