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Someone have me this problem claiming that I only need to use logs to solve it, though it seems to have a physics origin:

If a coil spring is stretched 1.5 m beyond its resisting point and then released it will return to a point which is 90 of the previous distance from the resting point. How many vibrations are required before the spring moves less tan 10 cm from its resting point?

I do not have much of a physics background, but could someone explain to me how this converts into a log equation? I tried myself several avenues (mainly of the form a = blog(c)^d) but nothing seems to work

A. Radek Martinez
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  • "90" means $90 %$ I guess ? So we have that the state of the spring at time $t=0$ is $150$ cm, let's write that $a_0 = 150$. We also know that $a_t = 0.9a_{t-1}$. Which means $a_t = 150 \cdot 0.9^t$. And you look for $a_t < 10$. – Zubzub Nov 24 '17 at 12:40
  • yes sorry I mean't 90%, If you write a solution I would happy to give you credit. – A. Radek Martinez Nov 24 '17 at 12:46
  • I already gave you everything you need. Just solve $150\cdot 0.9^t < 10$. If you ask such question you should be able to solve this inequality. (Divides by $150$ and then take the $\log$, then divides again (Be careful of the change of inequality sign)) – Zubzub Nov 24 '17 at 12:47
  • No no. I know you did, but I can not accept it, thus giving you points in this comment system. If you copy your answer over to "Answer Question" I can accept it. – A. Radek Martinez Nov 24 '17 at 12:49
  • Actually that's not exactly right, we have $a_t = (-1)^t 0.9a_{t-1}$ since it oscillate on both "sides" of the resting point. – Zubzub Nov 24 '17 at 12:53

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Let's think about discrete time step $t$ and let $x_t$ be the position of the end of the spring on the horizontal $x$-axis after $t$ oscillation. We suppose the resting point to be at $x=0$.

At start we have $x_0 = 150$, then after one osillation, it will go on the other side of the resting point and loose a bit of strength, that is $x_1 = -0.9 \cdot 150$, then go on the other side again: $x_2 = 150 \cdot 0.9 \cdot 0.9$.

We have the recurrence relation $x_t = -0.9 x_{t-1}, \ x_0 = 150$. This can be solve easily $x_t = (-1)^t 150\cdot 0.9^t$.

Now if we only care about "being less than 10cm from the resting point", we can consider only the absolute value of $x_t$ and get rid of the alternating sign.

Hence we look for $$ 150\cdot 0.9^t \leq 10 \implies 0.9^t \leq \frac{1}{15} \implies t\log(0.9) \leq -\log(15) $$ I let you finish. (And remember that $\log(0.9) < 0$, this is important when dealing with inequalities.)

Zubzub
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