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 November 5th, 2019, 04:07 AM #1 Senior Member     Joined: Jun 2017 From: Lima, Peru Posts: 178 Thanks: 5 Math Focus: Calculus How do I find a constant in an oscillation as given by a sphere in motion? 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?.
 November 5th, 2019, 05:47 AM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,093 Thanks: 1675 $s$ is displacement from equilibrium $a=-ks$ $\dfrac{dv}{dt} = -ks$ $\dfrac{dv}{dt} \cdot \dfrac{dt}{ds} = -\dfrac{ks}{v}$ $\dfrac{dv}{ds} = -\dfrac{ks}{v}$ $v \, dv = -ks \, ds$ $\dfrac{v^2}{2} = -\dfrac{ks^2}{2} + C_1$ $v^2 = -ks^2 + C_2$ when $s=5$, $v=0$ $\implies C_2=25k$ $v^2 = k(25-s^2)$ when $s=0$, $v=10 \implies k=4$ Had the mass of the sphere been given, an easier solution path could be found using conservation of energy. Thanks from topsquark and Chemist116 Last edited by skeeter; November 5th, 2019 at 06:31 AM.
 November 5th, 2019, 07:31 AM #3 Math Team     Joined: Jul 2011 From: Texas Posts: 3,093 Thanks: 1675 One final note ... the author of this problem used the equation $a = -ks$, which was a bit misleading at first. The actual equation should be $F = -ks \implies ma = -ks \implies a = -\dfrac{k}{m} \cdot s$ $F$ is the restoring force when an object is displaced from equilibrium, acting opposite in direction to the object's displacement. ... he/she just replaced the constant $\dfrac{k}{m}$ with a $k$ Thanks from topsquark and Chemist116
November 7th, 2019, 12:53 AM   #4
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Quote:
 Originally Posted by skeeter One final note ... the author of this problem used the equation $a = -ks$, which was a bit misleading at first. The actual equation should be $F = -ks \implies ma = -ks \implies a = -\dfrac{k}{m} \cdot s$ $F$ is the restoring force when an object is displaced from equilibrium, acting opposite in direction to the object's displacement. ... he/she just replaced the constant $\dfrac{k}{m}$ with a $k$
This was apparently the case. I noticed that for this problem to be solved you had to use the chain rule. This part was where I got an error in my attempt to integrate.

From the given conditions I could not "simplify" the solution as the mass was unknown. But if it had been known what steps should I could follow by conservation of energy?

Do you mean this energy?

$E_{p}=\frac{1}{2}kx^2$

Btw there's another problem (about rotational motion) which is unanswered I couldn't reach to the answer, would you mind taking a look into it?. Please

Last edited by Chemist116; November 7th, 2019 at 12:56 AM.

November 7th, 2019, 07:28 AM   #5
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Quote:
 Originally Posted by Chemist116 This was apparently the case. I noticed that for this problem to be solved you had to use the chain rule. This part was where I got an error in my attempt to integrate. From the given conditions I could not "simplify" the solution as the mass was unknown. But if it had been known what steps should I could follow by conservation of energy? Do you mean this energy? $E_{p}=\frac{1}{2}kx^2$
yes, also $K = \dfrac{1}{2}mv^2$

at equilibrium, $v = \dfrac{1}{10} \, m/s \implies K = E_{total} = \dfrac{m}{200} \, J$

at max displacement, $s=5$ (units not stated, maybe cm?)

$E_{total} = U = \dfrac{1}{2}ks^2$

solve for $k$

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