Question

Given that \[y = A\sin [(\dfrac{{2\pi }}{\lambda }\left( {ct - x} \right))]\], where y and x are measured in metres. Which of the following statements is true?
(A) The unit of λ is same as that of x and A
(B) The unit of λ is same as that of x but not of A
(C) The unit of c is same as that of $\dfrac{{2\pi }}{\lambda }$
(D) The unit of (ct−x) is same as that of $\dfrac{{2\pi }}{\lambda }$

Answer 1

Hint: The very aspect in order to deal with such a question is that the sine or cosine function have angles in the form of radian, and radians are unitless in their dimension form. So, we have to make the total value of the unit less in order to solve the problem. We use this key concept to eliminate the options here.

Complete answer:
Since our main concern is to make the function that are present inside the sine function dimensionless
$2\pi $ is a radian angle so it is dimensionless.
‘$x$’ has a unit of meters and dimension of length which means the dimension of ‘$ct$’ will also be the same as that of ‘$x$’ for addition and subtraction. Thus, ‘$ct$’ and $\left( {ct - x} \right)$ both have the dimension of length.
In order to make \[\dfrac{{2\pi }}{\lambda }\left( {ct - x} \right)\] the constant λ must have the dimension of length so that the dimension of $\left( {ct - x} \right)$ gets cancelled.
Now the sine function only gives a constant value without having any dimension to it (dimensionless value) so the dimension of ‘y’ will be the same as that of the dimension of A. Thus, the dimension of A is length.
Finally, we have $x$, $y$, λ and A all with dimensions of length.
Now the dimension of will be the same for speed as has been multiplied with time in order to get a dimension of length.
The dimension of $\dfrac{1}{\lambda }$ is the length inverse.

Note:
In this case we are given to deal with length problems inside a sine function. But in some cases, a time (frequency) variable is present or both time and length variables are present inside the sine function. In such cases we also have to make the time variables unit less in order to solve the problem. This is to make sure that the total variable inside the sine function sum up to be a dimensionless quantity.