A meter stick in frame ${\mathit{S}}^{\mathbf{\text{'}}}$ makes an angle of ${\mathbf{30}}^{\mathbf{o}}$with the x’ axis. If that frame moves parallel to the x-axis of frame $\mathit{S}$with speed 0.90c relative to frame $\mathit{S}$, what is the length of the stick as measured from S ?

The speed of frame relative to frame $S$ is $v=0.90c$
.

The length of the meter stick is $L_0=1m$$L_0=1m$.

The angle of the frame with the x-axis is $\theta ={30}^o$ .

The length contraction is used to find the length measured by the observer for the tube. The length of the meter stick only contracts in the x direction and the length of the meter stick remains the same. The length of the meter stick in frame is determined by the vector sum of the horizontal and vertical components of the length of the stick.

The length of the meter stick after contraction along the x-axis is given as:

$L_x=L_0\sqrt{1-{\left(\frac{v}{c}\right)}^2}$

Here, $c$ is the speed of light and its value is $3\times {10}^{8}m/s$.

Substitute all the values in the equation.

$$\begin{gathered} L_x=\left(1m\right)\sqrt{1-{\left(\frac{0.90}{c}\right)}^2} \\ {L}_x=0.436m \end{gathered}$$

The length of the meter stick in frame S is given as:

$L=\sqrt{{\left(L_x\cos \theta \right)}^2+{\left(L\sin \theta \right)}^2}$

Substitute all the values in the equation.

$$\begin{gathered} L=\sqrt{{\left[\left(0.436m\right)\left(\cos {30}^o\right)\right]}^2+{\left(L\sin \theta \right)}^2} \\ L=\sqrt{{\left[\left(0.436m\right)\left(0.866\right)\right]}^2+{\left[\left(1m\right)\left(0.5\right)\right]}^2} \\ L=\sqrt{0.1426+0.25} \\ L=0.63m \end{gathered}$$

Hence, the length of meter stick in frame$S$ is 0.63 m .