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borishaifa [10]

3 years ago

9

A negatively charged particle -q is placed at the center of a uniformly charged ring, where the ring has a total positive charge

Q as shown in Example 19.5. The particle, confined to move along the x axis, is displaced a small distance x along the axis (x is much smaller than a) and released. Show that the particle oscillates in simple harmonic and calculate its frequency.

1 answer:

navik [9.2K]3 years ago

5 0

The solution would be like this for this specific problem:

<span>
F=−</span>k∗x∗<span>q∗</span>Q<span>/(</span>+)<span>F−≈</span><span><span><span>k∗x∗<span>q∗</span>Q</span><span>/R3</span></span>[(</span>1−<span><span>3/2</span><span><span>*x2</span><span>/R3</span></span>]
</span><span>F=−</span><span><span>k∗x∗<span>q∗/</span>Q</span><span>R<span>3
</span></span></span><span>F=</span><span>ma
</span>−<span><span><span><span>k∗<span>q∗</span>Q</span><span>/R3</span></span>*</span>x</span>=<span>ma
</span>−k∗x=m∗<span>a
a</span>==<span><span><span>ω2</span>x
</span>ω</span><span>=(</span>k/<span>m<span>)<span><span>1/</span><span>2

</span></span></span></span>ω<span>=(</span><span>kqQ</span>/<span><span>R3</span><span>)<span><span>1/</span>2

</span></span></span>

<span>I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.</span>

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The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

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Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

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$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

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