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We have the following equation $$mg\sin(\theta)=kx\left(1-\frac{L}{\sqrt{x^2+a^2}}\right)$$ and we are asked to put it in the following dimensionless form: $$1-\frac{h}{u}=\frac{R}{\sqrt{1+u^2}}.$$ According to me $u=x/a$ and $R=L/a$ then we have $$\frac{R}{\sqrt{1+u^2}}=1-\frac{mg\sin(\theta)}{kx}$$ which would imply that $$\frac{h}{u}=\frac{mg\sin(\theta)}{kx}$$ which is weird since $u=x/a.$ Where am I going wrong?

Edit: Here I am adding the units of each term in the equation given above:

$m:\text{mass},$ $g:\text{acceleration due to gravity so }ms^{-2}$,$k:\text{ spring constant so }Nm^{-1},\{x,L,a\}:\text{length}.$

Here is the original question Part(b) in case I made any typos:

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Student
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1 Answers1

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Let us rewrite the equilibrium position of the bead as \begin{align*} \frac{mg\sin\theta}{kx} & = 1 - \frac{L_0}{\sqrt{x^2 + a^2}} \\ 1 - \frac{mg\sin\theta}{kx} & = \frac{L_0}{\sqrt{x^2 + a^2}} \\ 1 - \frac{mg\sin\theta}{kx} & = \frac{L_0}{a\sqrt{(x/a)^2 + 1}} \end{align*} Define the dimensionless variable $u = x/a$. Then \begin{align*} 1 - \frac{mg\sin\theta}{kau} & = \frac{L_0}{a\sqrt{u^2 + 1}}. \end{align*} The corresponding $R$ and $h$ are $$ R = \frac{L_0}{a} \ \ \textrm{ and } \ \ h = \frac{mg\sin\theta}{ka}. $$ Let us check that $R$ and $h$ are indeed dimensionless. It is clear that $R$ is dimensionless since $L_0$ and $a$ are lengths. The dimension of $mg$ is $$ [mg] = \frac{\textrm{mass}\times\textrm{length}}{\textrm{time}^2} $$ and the dimension of $k$ is $$ [k] = \frac{[mg]}{[x]} = \frac{\textrm{mass}}{\textrm{time}^2}. $$ Consequently, $h$ is also dimensionless. Here, the square bracket denotes the dimension of the quantity of interest.

Chee Han
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