The frequency f of vibration of mass m suspended from a spring of spring contact k is given by f-c -m-x- -k-y- Where c- dimensionless constant- then the values of x and y-

Question

Toppr Admin · Accepted Answer

In this situation dimension of RHS-LHSi-e- -xA0- -f-cmxky-xA0- -xA0- -xA0- -xA0-T-x2212-1-Mx-MyT-x2212-2y-Mx-yT-x2212-2y- -xA0- since -k-MT-x2212-2-comparing both sides we get-xA0-xA0- -xA0- -xA0- -xA0-x-y-0-and-x2212-1-x2212-2y-x21D2-y-1-2-xA0- -xA0- -xA0-and -xA0- x-x2212-y-x2212-1-2

The frequency $f$ of vibration of mass $m$ suspended from a spring of spring contact $k$ is given by $f = c m^{x} k^{y}$ . Where $c =$ dimensionless constant, then find the values of $x$ and $y$ .

In this situation dimension of RHS=LHS
i.e, $[f] = [c m^{x} k^{y}]$
$[T^{- 1}] = [M^{x}] [M^{y} T^{- 2 y}] = [M^{x + y} T^{- 2 y}]$ , since $[k] = [M T^{- 2}]$
comparing both sides we get,
$x + y = 0, a n d - 1 = - 2 y \Rightarrow y = 1 / 2$
and $x = - y = - 1 / 2$

The frequency f of vibration of mass m suspended from a spring of spring contact k is given by f=c mx ky. Where c= dimensionless constant, then find the values of x and y.

In this situation dimension of RHS=LHSi.e, [f]=[cmxky] [T−1]=[Mx][MyT−2y]=[Mx+yT−2y], since [k]=[MT−2]comparing both sides we get, x+y=0,and−1=−2y⇒y=1/2 and x=−y=−1/2

The frequency $f$ of vibration of mass $m$ suspended from a spring of spring contact $k$ is given by $f = c m^{x} k^{y}$ . Where $c =$ dimensionless constant, then find the values of $x$ and $y$ .

In this situation dimension of RHS=LHS
i.e, $[f] = [c m^{x} k^{y}]$
$[T^{- 1}] = [M^{x}] [M^{y} T^{- 2 y}] = [M^{x + y} T^{- 2 y}]$ , since $[k] = [M T^{- 2}]$
comparing both sides we get,
$x + y = 0, a n d - 1 = - 2 y \Rightarrow y = 1 / 2$
and $x = - y = - 1 / 2$