Question

1. (5pts) using the fact that the argument of the cosine function must be dimensionless, find the units of ( a ) if the position of an object is described by x(t) = x_0 cosleft( t sqrt{a}

ight) where ( x ) is the object’s velocity and ( t ) is the time

Explanation:

Step1: Analyze the argument of cosine

The argument of the cosine function, \( t\sqrt{a} \), must be dimensionless (since trigonometric functions take dimensionless arguments). Let the units of \( t \) be time (\( T \)) and the units of \( a \) be \( [a] \). So, the units of \( t\sqrt{a} \) should be dimensionless, which means \( [t] \times \sqrt{[a]} = 1 \) (dimensionless).

Step2: Solve for the units of \( a \)

We know \( [t] = T \) (time units). So, \( T \times \sqrt{[a]} = 1 \) (dimensionless). Rearranging for \( \sqrt{[a]} \), we get \( \sqrt{[a]}=\frac{1}{T} \). Squaring both sides, \( [a]=\frac{1}{T^{2}} \). In terms of physical quantities, acceleration has units of length per time squared, but here we are dealing with the argument's dimensional analysis. Wait, actually, if we consider the position function \( x(t) = x_0\cos(t\sqrt{a}) \), the argument \( t\sqrt{a} \) must be dimensionless. Let's assume \( x \) has units of length (\( L \)), but the cosine is a dimensionless function. The key is the argument: \( t \) is time (\( T \)), so \( \sqrt{a} \) must have units of \( 1/T \) (so that \( T \times 1/T = 1 \), dimensionless). Therefore, \( a \) has units of \( 1/T^{2} \). But if we think about the standard simple harmonic motion equation, \( x(t) = A\cos(\omega t) \), where \( \omega=\sqrt{k/m} \) (for spring - mass) or \( \omega=\sqrt{g/L} \) (for pendulum), and the units of \( \omega \) are \( 1/T \) (angular frequency). So comparing \( \omega = \sqrt{a} \), then \( a \) would have units of \( \omega^{2} \), i.e., \( 1/T^{2} \). But if we consider the physical meaning of the position function, if \( x \) is position (length, \( L \)), and we know from the equation of motion for simple harmonic motion \( \ddot{x}=-a x \) (second derivative of \( x \) with respect to \( t \) is \( -a x \)). The second derivative of position (length) with respect to time (time) has units of \( L/T^{2} \), and the right - hand side is \( -a x \), with \( x \) having units of \( L \). So \( [\ddot{x}]=L/T^{2} \) and \( [-a x]=[a]\times L \). Therefore, equating the units: \( L/T^{2}=[a]\times L \), so \( [a]=1/T^{2} \). But maybe in the context of the problem, we can also think in terms of the given function. Since the argument \( t\sqrt{a} \) is dimensionless, let \( [t] = T \), then \( \sqrt{[a]}=\frac{1}{T} \), so \( [a]=\frac{1}{T^{2}} \). If we consider the units of velocity \( v \) (from the note "where \( v \) is the object's velocity" - wait, maybe there was a typo, and it's supposed to be \( \omega=\sqrt{a} \), and velocity \( v \) has units of \( L/T \), but maybe that's a distraction. The main thing is the dimensional analysis of the cosine argument. So, from \( t\sqrt{a} \) being dimensionless, \( [t]=T \), so \( \sqrt{[a]} = 1/T\implies [a]=1/T^{2} \). But if we assume that the position \( x \) has units of length (\( L \)), and we know that the second derivative of \( x \) with respect to \( t \) (acceleration) has units of \( L/T^{2} \). In the equation \( \ddot{x}=-a x \), \( [\ddot{x}]=L/T^{2} \) and \( [-a x]=[a]\times L \). So \( L/T^{2}=[a]\times L\implies [a]=1/T^{2} \). But if we are just doing dimensional analysis on the argument of the cosine:
Let \( u = t\sqrt{a} \), \( [u]=1 \) (dimensionless). \( [t]=T \), so \( [\sqrt{a}]=\frac{[u]}{[t]}=\frac{1}{T} \). Then \( [a]=(\frac{1}{T})^{2}=\frac{1}{T^{2}} \). If we consider the physical quantity, acceleration has units of \( L/T^{2} \), but in the context of the cosine argument's dimensional analysis, we are only concerned with the time - related part here. Wait, maybe the problem is simpler. The argument of the cosine must be dimensionless. So \( t\sqrt{a} \) is dimensionless. Let \( t \) have units of time (say, seconds, \( s \)) and \( a \) have units \( [a] \). Then \( s\times\sqrt{[a]} \) must be dimensionless. So \( \sqrt{[a]}=\frac{1}{s} \), so \( [a]=\frac{1}{s^{2}} \). So the units of \( a \) are \( 1/\text{time}^2 \). If we think in terms of the standard simple harmonic motion, where the angular frequency \( \omega=\sqrt{a} \), and \( \omega \) has units of \( \text{rad/s} \) (but radians are dimensionless), so \( \omega \) effectively has units of \( 1/s \), so \( a=\omega^{2} \) has units of \( 1/s^{2} \).

Answer:

The units of \( a \) are \( \frac{1}{\text{time}^2} \) (or in SI units, \( \text{s}^{-2} \)). If we consider the physical context of simple harmonic motion, and relate it to acceleration, if the position function is similar to simple harmonic motion, \( a \) would have units of acceleration (length per time squared) in the equation of motion \( \ddot{x}=-a x \), but from the dimensional analysis of the cosine argument, we find that the units of \( a \) are \( \frac{1}{T^{2}} \) (where \( T \) is the unit of time).