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: \(x=2\).
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,


What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)