How do I find a constant in an oscillation as given by a sphere in motion?

Question

The problem is as follows:

The acceleration of an oscillating sphere is defined by the equation $a=-ks$. Find the value of $k$ such as $v=10\,\frac{cm}{s}$ when $s=0$ and $s=5$ when $v=0$.

The given alternatives in my book are as follows:

$\begin{array}{ll} 1.&15\\ 2.&20\\ 3.&10\\ 4.&4\\ 5.&6\\ \end{array}$

What I attempted to do here is to use integration to find the value of $k$.

Since the acceleration measures the rate of change between the speed and time then:

I'm assuming that they are using a weird notation "s" for the time.

$\dfrac{d(v(s))}{ds}=-ks$

$v(s)=-k\frac{s^2}{2}+c$

Using the given condition: $v(0)=10$

$10=c$

$v(s)=-k\frac{s^2}{2}+10$

Then it mentions: $v(5)=0$

$0=-k\frac{25}{2}+10$

From this it can be obtained:

$k=\frac{20}{25}=\frac{4}{5}$

However this value doesn't appear within the alternatives. What part did I missunderstood?. Can somebody help me here?.

Answer 1

The term oscillating is very generic, so this could be a pendulum for example. I'm assuming $s$ stands for the position of the sphere. Then, notice that $$a =\frac{dv}{dt} = \frac{dv}{ds}\frac{ds}{dt} = \frac{dv}{ds}v$$ Therefore, we have $$\int v dv = -\int ks ds \Rightarrow \frac{1}{2}{v}^{2}= -\frac{1}{2}ks^{2}+C$$ Now, if $s=0$ we have $v=10$ (I'm ignoring units) so $C = 50$. Now, if $s = 5$, $v = 0$ and this implies $k=4$.

Answer 2

The first thing to realize is that $s$ is not a "weird notation for time". It's a fairly standard notation for distance along a path. It is used much the same way as $x$ is used in other problems, except that we tend to use $x$ when we know the motion is in a straight line and can declare that line to be the $x$ axis, whereas not every oscillation problem involves straight-line motion.

So $v = \frac{ds}{dt}$ and $a = \frac{dv}{dt}.$

Then you might recognize $a = -ks$ as an example of Hooke's Law and you might recall that the result is sinusoidal oscillation around $s = 0.$ But you don't actually need to know that.

All you really need to know is that along the path of length $5$ units from $s = 0$ to $s = 5,$ the velocity of the sphere is reduced from $v = 10$ to $v = 0.$ If you let the letter $m$ stand for the mass of the sphere, then the force $F = ma = -mks$ acts on the sphere along that path. You can integrate over $s$ to figure out how much work that force does on the sphere and match that to the change in kinetic energy of the sphere. The change is negative, because we go from whatever kinetic energy the sphere has at $v = 10$ (I hope you remember the formula for that) to zero at $v = 0$.