The frequency of vibration f of a mass m suspended from a spring of spring constant k is given by a relation of the type f=c mx ky, where c is a dimensionless constant. The values of x and y are

Question

The frequency of vibration f of a mass m suspended from a spring of spring constant k is given by a relation of the type f=c mx ky, where c is a dimensionless constant. The values of x and y are (A) x=(1/2), y=(1/2) (B) x= -(1/2), y= -(1/2) (C) x=(1/2), y= -(1/2) (D) x= -(1/2), y=(1/2)

Tardigrade Admin · Accepted Answer

f=c mx ky Spring constant k= force/length. [M0 L0 T-1]=[Mx((M T)-2)y]=[Mx+y T-2 y] ⇒ x+y=0,-2 y=-1 or y=(1/2) Therefore, x=-(1/2)

Q. The frequency of vibration $f$ of a mass m suspended from a spring of spring constant $k$ is given by a relation of the type $f = c m^{x} k^{y},$ where $c$ is a dimensionless constant. The values of $x$ and $y$ are

$x = \frac{1}{2}, y = \frac{1}{2}$

$x = - \frac{1}{2}, y = - \frac{1}{2}$

$x = \frac{1}{2}, y = - \frac{1}{2}$

$x = - \frac{1}{2}, y = \frac{1}{2}$

$f = c m^{x} k^{y}$
Spring constant $k =$ force/length.
$[M^{0} L^{0} T^{- 1}] = [M^{x} ((MT)^{- 2})^{y}] = [M^{x + y} T^{- 2 y}]$
$\Rightarrow x + y = 0, - 2 y = - 1$ or $y = \frac{1}{2}$
Therefore, $x = - \frac{1}{2}$

Q. The frequency of vibration f of a mass m suspended from a spring of spring constant k is given by a relation of the type f=cmxky, where c is a dimensionless constant. The values of x and y are

x=21​,y=21​

x=−21​,y=−21​

x=21​,y=−21​

x=−21​,y=21​