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November 5th, 2019, 04:07 AM   #1
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Question 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?.
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November 5th, 2019, 05:47 AM   #2
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$s$ 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.
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Last edited by skeeter; November 5th, 2019 at 06:31 AM.
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November 5th, 2019, 07:31 AM   #3
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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 $k$
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November 7th, 2019, 12:53 AM   #4
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Math Focus: Calculus
Question

Quote:
Originally Posted by skeeter View Post
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.
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November 7th, 2019, 07:28 AM   #5
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Quote:
Originally Posted by Chemist116 View Post
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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