Question

In the expression $y = a\sin (\omega t + \theta )$, $y$ is the displacement and $t$ is the time. Write the dimensions of $\omega$.
A. $\left[ {{M^0}{L^1}{T^0}} \right]$
B. \[\left[ {{M^0}{L^0}{T^0}} \right]\]
C. \[\left[ {{M^0}{L^0}{T^{ - 1}}} \right]\]
D. \[\left[ {{M^1}{L^0}{T^0}} \right]\]

Answer 1

Hint: The argument of a trigonometric function that is angle a dimensionless quality. So compare the dimensions of both sides.
Dimension of $\omega$t = Dimension of $\theta$

Complete step by step solution:
The quantities which have the same dimensions only can be added or subtracted. So, $\omega$t and $\theta$ must have the same dimensions because they are added.
Dimension of $t = \left[ {{M^0}{L^0}{T^1}} \right]$
Dimension of $\theta = \left[ {{M^0}{L^0}{T^0}} \right]$
Where $\left[ M \right]$the dimension of mass is, \[\left[ L \right]\] is the dimension of length & \[\left[ T \right]\] is the dimension of time.
So, dimension of $\omega$t = dimension of $\theta$
dimension of $\omega$ = dimension of $\theta$ /dimension of t
dimension of $\omega$\[ = \dfrac{{\left[ {{M^0}{L^0}{T^0}} \right]}}{{\left[ {{M^0}{L^0}{T^1}} \right]}}\]
When the dimension goes from denominator to numerator sign of power on dimension has inverted.
$\Rightarrow$$\omega$\[ = \left[ {{M^0}{L^0}{T^0}} \right]\left[ {{T^{ - 1}}} \right]\]
On multiplication and division, dimensions follow operations similar to logarithms.
$\Rightarrow$$\omega$$ = \left[ {{M^0}{L^0}{T^{0 - 1}}} \right]$
$\Rightarrow$$\omega$$ = \left[ {{M^0}{L^0}{T^{ - 1}}} \right]$

$\therefore$ The dimensions of $\omega$ is $\left[ {{M^0}{L^0}{T^{ - 1}}} \right]$. Hence, option (C) is correct.

Note:
Maximum time’s student thought that $\theta$ is dimensionless. So, they can’t compare $\omega t$ with $\theta$ and start a comparison between $\omega t$ and $y$, which is the wrong approach.