The equation of a stationary wave is given by | vedclass

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

Question

(C) In the argument of the sine function,$\frac{2 \pi nt}{\lambda}$ must be dimensionless. Since $2\pi$ is dimensionless,the dimensions of $\frac{nt}{

Vedclass · Accepted Answer

The dimensions of $n/\lambda$ are $[T^{-1}]$.

Vedclass · Answer

(C) In the argument of the sine function,$\frac{2 \pi nt}{\lambda}$ must be dimensionless. Since $2\pi$ is dimensionless,the dimensions of $\frac{nt}{\lambda}$ must be $[M^0 L^0 T^0]$.This implies $[nt] = [\lambda] = [L]$. Thus,option $A$ is correct.Given $[nt] = [L]$ and $[t] = [T]$,we have $[n] = [L/T] = [LT^{-1}]$. Thus,option $B$ is correct.For the cosine function,$\frac{2 \pi x}{\lambda}$ must be dimensionless,so $[x] = [\lambda] = [L]$. Thus,option $D$ is correct.Now,consider the dimensions of $n/\lambda$. Since $[n] = [LT^{-1}]$ and $[\lambda] = [L]$,we have $[n/\lambda] = [LT^{-1}] / [L] = [T^{-1}]$.Option $C$ states the dimensions are $[T]$,which is incorrect. Therefore,option $C$ is the correct answer.

The equation of a stationary wave is given by $y = 2a \sin \left( \frac{2 \pi nt}{\lambda} \right) \cos \left( \frac{2 \pi x}{\lambda} \right)$. Which of the following statements is $NOT$ correct?

Similar Questions

In an experiment,four quantities $a, b, c$ and $d$ are measured with percentage errors of $1\%, 2\%, 3\%$ and $4\%$ respectively. Quantity $w$ is calculated as follows: $w = \frac{a^4 b^3}{c^2 \sqrt{d}}$. The percentage error in the measurement of $w$ is .......... $\%$.

The potential energy of a particle varies with distance $x$ from a fixed origin as $U = \frac{A \sqrt{x}}{x^2 + B}$,where $A$ and $B$ are dimensional constants. Then,the dimensional formula for $AB$ is:

The velocity $v$ (in $cm/sec$) of a particle is given in terms of time $t$ (in $sec$) by the relation $v = at + \frac{b}{t + c}$. The dimensions of $a, b,$ and $c$ are:

In a typical combustion engine,the work done by a gas molecule is given by $W = \alpha^{2} \beta e^{\frac{-\beta x^{2}}{kT}}$,where $x$ is the displacement,$k$ is the Boltzmann constant,and $T$ is the temperature. If $\alpha$ and $\beta$ are constants,the dimensions of $\alpha$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module