The equation of the stationary wave is y = 2A | vedclass

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

Question

(D) In the equation $y = 2A \sin \left( \frac{2\pi ct}{\lambda} \right) \cos \left( \frac{2\pi x}{\lambda} \right)$,the arguments of the trigonometric

Vedclass · Accepted Answer

The unit of $c/\lambda$ is same as that of $x/\lambda$

Vedclass · Answer

(D) In the equation $y = 2A \sin \left( \frac{2\pi ct}{\lambda} ight) \cos \left( \frac{2\pi x}{\lambda} ight)$,the arguments of the trigonometric functions must be dimensionless.Therefore,$\left[ \frac{2\pi ct}{\lambda} ight] = M^0 L^0 T^0$ and $\left[ \frac{2\pi x}{\lambda} ight] = M^0 L^0 T^0$.From this,we conclude that the units of $ct$ and $\lambda$ are the same,and the units of $x$ and $\lambda$ are the same.Checking option $C$: The unit of $\frac{2\pi c}{\lambda}$ is $T^{-1}$ (frequency),while the unit of $\frac{2\pi x}{\lambda t}$ is $\frac{L}{L \cdot T} = T^{-1}$. Thus,they have the same units.Checking option $D$: The unit of $\frac{c}{\lambda}$ is $T^{-1}$,while the unit of $\frac{x}{\lambda}$ is dimensionless $(L/L = 1)$. Since $T^{-1} 
eq 1$,statement $D$ is not true.

The equation of the stationary wave is $y = 2A \sin \left( \frac{2\pi ct}{\lambda} \right) \cos \left( \frac{2\pi x}{\lambda} \right)$. Which statement is not true?

Similar Questions

If force $(F)$,velocity $(V)$,and time $(T)$ are taken as fundamental units,then the dimensions of mass are:

The equation $\frac{dV}{dt} = At - BV$ describes the rate of change of velocity of a body falling from rest in a resisting medium. The dimensions of $A$ and $B$ are

If the charge of electron $e$,mass of electron $m$,speed of light in vacuum $c$,and Planck's constant $h$ are taken as fundamental quantities,then the permeability of vacuum $\mu_0$ can be expressed as

If $C$ (the velocity of light), $h$ (Planck's constant), and $G$ (gravitational constant) are taken as fundamental quantities, then the dimensional formula of mass is

In the expression $A=B+\frac{C}{D+E}$,the dimensions of physical quantities $B$ and $C$ are $[L^{1} M^{0} T^{-1}]$ and $[L^{1} M^{0} T^{0}]$ respectively. The dimensions of quantities $A, D$ and $E$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module