The equation of motion of a damped oscillator | vedclass

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(A) The given equation is $m \frac{d^2 x}{dt^2} + b \frac{dx}{dt} + kx = 0$. According to the principle of homogeneity of dimensions, each term in the

Vedclass · Accepted Answer

$[M^0 L^0 T^0]$

Vedclass · Answer

(A) The given equation is $m \frac{d^2 x}{dt^2} + b \frac{dx}{dt} + kx = 0$. According to the principle of homogeneity of dimensions, each term in the equation must have the same dimensions. Thus, the dimensions of $m \frac{d^2 x}{dt^2}$, $b \frac{dx}{dt}$, and $kx$ are equal. Equating the dimensions of $b \frac{dx}{dt}$ and $kx$: $[b] [v] = [k] [x] \implies [b] = [k] [x] / [v] = [k] [x] / ([x] / [t]) = [k] [t]$. Therefore, $[b] = [k] [T]$. Now, consider the expression $\frac{b}{\sqrt{km}}$. Substituting $[b] = [k] [T]$ into the expression: $\frac{[b]}{\sqrt{[k][m]}} = \frac{[k][T]}{\sqrt{[k][m]}}$. Since $k = F/x = ma/x$, the dimensions of $k$ are $[M T^{-2}]$. Also, the angular frequency of a damped oscillator is related to $\omega = \sqrt{k/m}$, so $\sqrt{k/m}$ has dimensions of $[T^{-1}]$. Alternatively, using $[b] = [k] [T]$: $\frac{b}{\sqrt{km}} = \frac{k T}{\sqrt{km}} = \sqrt{\frac{k}{m}} T = [T^{-1}] [T] = [M^0 L^0 T^0]$.

The equation of motion of a damped oscillator is given by $m \frac{d^2 x}{d t^2}+b \frac{d x}{d t}+k x=0$. The dimensional formula of $\frac{b}{\sqrt{k m}}$ is

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