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We have a coil spring such that a $25\,\textbf{lb}$ weight it will stretched a length of $6\,\textbf{in}$. A mass of weight $16\,\textbf{lb}$ is attached to the spring. The string automatically stretched and in a rest position now. The the mass is stretched further $4\,\textbf{in}$. The object is released with an initial velocity of $2\,\textbf{ft}/\textbf{sec}$ that is directed upward.

1) Determine the resulting displacement as a function of time.

$$m\frac{d^2x}{dt^2}+kx=0 \\ m=\frac {16}{32}=\frac12 ~, \quad mg=kl \\ 25=k\frac{6}{12} ~,\quad k=50 \\ \implies \frac{1}{2}\frac{d^2x}{dt^2}+50x=0$$

Auxiliary equations:

$$ y = c_1 \sin(10t) + c_2 \cos (10t) \\ c_2=\frac{1}{3} \quad c_1=-\frac{1}{5}\\ \implies y=\frac{-1}{5}\sin 10t+\frac{1}{3}\cos 10t \\ \implies y=\frac{\sqrt 34}{15}\cos (10t+0.5404) $$

  1. Determine the amplitude, period and frequency of the mass.

$$A=\frac{\sqrt 34}{15}~, \quad T=\frac{\pi}{5}~, \quad f=\frac{5}{\pi}$$

  1. What is the time when the weight first pass through the equilibrium position what is the time at this instant.

My work is

$$0=\cos (10t+0.5404)$$

$$t=0.1030\, \textbf{s} $$

By my intuition

$$ T=\frac{\pi}{5}$$

A quarter time is

$$t=\frac{\pi}{20} \approx 0.157\, \textbf{s}$$

What I want to ask if which answer shall I accept for the third part? There is no back answer for this question. I hope that someone guide me to the right way. Thanks in advance.

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By setting up the ODE in the manner you did, you have established that the equilbirium position is at x = 0. Therefore, I would say that $t = 0.1030$ , if your calculations are correct, is the right answer.

The period does not have much with the equilibrium position. The period only tells you how long does it take for the mass to get to the same position at which it started. Therefore, assuming that, at a quarter of the period, the mass will be at zero is a an arbitrary assumption; which does not work here.

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