Question

The equation of motion of a damped oscillator is given by \( m\frac{d^{2}x}{dt^{2}}+b\frac{dx}{dt}+kx=0 \). The dimensional formula of \( \frac{b}{\sqrt{km}} \) is

• A. \([M^{0}L^{0}T^{0}]\)
• B. \([M^{0}L^{1}T^{-2}]\)
• C. \([M^{1}L^{1}T^{-2}]\)
• D. \([M^{1}L^{2}T^{-2}]\)

Hint:

we need to find the dimensional formula, first find the dimensions of each term in the equation.

Answer 1

The Correct Option is A

Given the equation \( m\frac{d^{2}x}{dt^{2}}+b\frac{dx}{dt}+kx=0 \), we need to find the dimensional formula of \( \frac{b}{\sqrt{km}} \). 
First, let's find the dimensions of each term in the equation.

• \( m\frac{d^{2}x}{dt^{2}} \): Mass \(\times\) acceleration, which has dimensions \([M][LT^{-2}] = [MLT^{-2}]\).
• \( b\frac{dx}{dt} \): \(b\) \(\times\) velocity, which has dimensions \([b][LT^{-1}]\).
• \( kx \): \(k\) \(\times\) displacement, which has dimensions \([k][L]\).

Since all terms must have the same dimensions, we have:

• \([b][LT^{-1}] = [MLT^{-2}] \implies [b] = [MLT^{-1}]\)
• \([k][L] = [MLT^{-2}] \implies [k] = [MT^{-2}]\)

Now, let's find the dimensions of \( \frac{b}{\sqrt{km}} \): 
\[ \left[ \frac{b}{\sqrt{km}} \right] = \frac{[MLT^{-1}]}{\sqrt{[MT^{-2}][M]}} = \frac{[MLT^{-1}]}{\sqrt{[M^2T^{-2}]}} = \frac{[MLT^{-1}]}{[MT^{-1}]} = [M^{1-1}L^{1-0}T^{-1-(-1)}] = [M^{0}L^{0}T^{0}] \] 
Final Answer: (1) \([M^{0}L^{0}T^{0}]\).