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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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Math Focus: Calculus 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 $k$ Tags constant, find, motion, oscillation, sphere Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Statistics132 Applied Math 4 November 5th, 2016 04:43 PM rhubarb123 Algebra 2 May 16th, 2014 10:38 PM WWRtelescoping Advanced Statistics 2 April 13th, 2014 10:51 AM OrangeLightning Elementary Math 6 April 23rd, 2013 03:14 PM gholamghar Applied Math 0 July 27th, 2009 12:35 AM

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