The frequency of vibration f of a mass m susp | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The dimensional formula for frequency $f$ is $[M^0 L^0 T^{-1}]$.The dimensional formula for mass $m$ is $[M^1 L^0 T^0]$.The dimensional formula fo

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The dimensional formula for frequency $f$ is $[M^0 L^0 T^{-1}]$.The dimensional formula for mass $m$ is $[M^1 L^0 T^0]$.The dimensional formula for spring constant $K$ is $[M^1 L^0 T^{-2}]$.Given the relation $f = C m^x K^y$,substituting the dimensions on both sides:$[M^0 L^0 T^{-1}] = [M^1]^x [M^1 T^{-2}]^y$$[M^0 L^0 T^{-1}] = [M^{x+y} T^{-2y}]$Comparing the powers of $M$ and $T$ on both sides:For $M$: $x + y = 0$For $T$: $-2y = -1 \implies y = \frac{1}{2}$Substituting $y = \frac{1}{2}$ into $x + y = 0$,we get $x + \frac{1}{2} = 0 \implies x = -\frac{1}{2}$.Thus,$x = -\frac{1}{2}$ and $y = \frac{1}{2}$.

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 quantity. The values of $x$ and $y$ are:

Similar Questions

The equation of a circle is given by $x^2+y^2=a^2$,where $a$ is the radius. If the equation is modified to change the origin to a point other than $(0,0)$,find the correct dimensions of $A$ and $B$ in the new equation: $(x-At)^2+(y-\frac{t}{B})^2=a^2$. The dimensions of $t$ are given as $[T^{-1}]$.

The dimensions of $\frac{\alpha}{\beta}$ in the equation $F = \frac{\alpha - t^2}{\beta v^2}$,where $F$ is force,$v$ is velocity,and $t$ is time,are ..........

The formula is $X = 5YZ^2$,where $X$ and $Z$ have dimensions of capacitance and magnetic field,respectively. What are the dimensions of $Y$ in $SI$ units?

Given that: $\lambda = a \cos \left( \frac{t}{p} - qx \right)$,where $t$ represents time in $s$ and $x$ represents distance in $m$. Which of the following statements is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module