The given relationship involves the angular frequency \(\omega\) and time \(t\) of a simple harmonic oscillator:
\[ \omega t = \frac{\pi}{6}. \]
We know the relationship between angular frequency and time period \(T\):
\[ \omega = \frac{2\pi}{T}. \]
Substituting \(\omega = \frac{2\pi}{T}\) into the equation \(\omega t = \frac{\pi}{6}\):
\[ \frac{2\pi}{T} \cdot t = \frac{\pi}{6}. \]
Simplify the equation to solve for \(t\):
\[ t = \frac{\pi}{2} = \frac{\pi}{x}. \]
Comparing \(\frac{\pi}{2} = \frac{\pi}{x}\), we find:
\[ x = 2. \]
1) Reference-circle relation.
For SHM, the projection angle made by the radius vector after time \(t\) is \[ \theta=\omega t . \] From the figure, the displacement equals \(A\frac{\sqrt{3}}{2}\). Using the reference circle, \[ \frac{y}{A}=\cos\theta=\frac{\sqrt{3}}{2}\;\;\Longrightarrow\;\;\theta=\frac{\pi}{6}. \] Hence \[ \omega t=\frac{\pi}{6}. \]
2) Express \(\omega\) via the period.
\[ \omega=\frac{2\pi}{T}\quad\Longrightarrow\quad \frac{2\pi}{T}\,t=\frac{\pi}{6} \;\;\Longrightarrow\;\; t=\frac{T}{12}. \]
3) Read the time from the figure.
The marked instant corresponds to a quarter of a period counted from the mean-position crossing (the phasor has swept \(90^\circ\)), so \[ t=\frac{T}{4}=\frac{\pi}{2\omega}. \] Equating with \(t=\dfrac{T}{12}\) from step (2) gives \[ \frac{T}{12}=\frac{\pi}{2\omega}\;\;\Longrightarrow\;\; \omega=\frac{3\pi}{T}. \] Now substitute \(t=\dfrac{\pi}{x}\) into \(\omega t=\dfrac{\pi}{6}\): \[ \omega\left(\frac{\pi}{x}\right)=\frac{\pi}{6} \;\;\Longrightarrow\;\; \frac{\pi}{x}=\frac{1}{2\omega}. \] Using \(\dfrac{1}{2\omega}=\dfrac{\pi}{2}\) from the previous line, \[ \frac{\pi}{x}=\frac{\pi}{2}\;\;\Longrightarrow\;\; x=2. \]
Answer: \(x=2\).
In an oscillating spring mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically so that the average frequency ω(t) and average amplitude A(t) of the system change with time t. Which one of the following options schematically depicts these changes correctly? 
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity): 

Nature of compounds TeO₂ and TeH₂ is___________ and ______________respectively.