First slide

Dimensions of physical quantities

Question

The frequency $f$ of vibrations of a mass m suspended from a spring of spring constant k is given by $f = C m^{x} k^{y}$ , where C is a dimensionless constant. The values of x and y are, respectively,

Moderate

A

$\frac{1}{2}, \frac{1}{2}$
B

$- \frac{1}{2}, - \frac{1}{2}$
C

$\frac{1}{2}, - \frac{1}{2}$
D

$- \frac{1}{2}, \frac{1}{2}$

Solution

force constant K= $\frac{F}{∆ x}$
$f = C m^{x} k^{y}$ . Writing dimensions on both sides,
$[M^{0} L^{0} T^{- 1}] = M^{x} {[M L^{0} T^{- 2}]}^{y} = [M^{x + y} T^{- 2 y}]$
Comparing dimensions on both sides, we have
$0 = x + y and - 1 = - 2 y \Rightarrow y = \frac{1}{2}, x = - \frac{1}{2}$
Aliter. Remembering that the frequency of oscillation of loaded spring is
$\begin{array}{l} f = \frac{1}{2 π} \sqrt{\frac{k}{m}} = \frac{1}{2 π} (k)^{1 / 2} m^{- 1 / 2} \\ which gives x = - \frac{1}{2} and y = \frac{1}{2} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App