The velocity component vPS(t) along the line joining the source and the observer is given by:
vPS(t) = v(t) cosΟ
where Ο is the angle between the velocity vector and the line joining S and P.
From the geometry of the problem, the distance between the source and the observer is:
r(t) = β(x(t)Β² + LΒ²) = β(aΒ² sinΒ²(Οt) + LΒ²)
The angle Ο satisfies:
cosΟ = L / r(t)
Substituting v(t) = v0 cos(Οt) and cosΟ = L / r(t), we have:
vPS(t) = v0 cos(Οt) Γ (L / β(aΒ² sinΒ²(Οt) + LΒ²))
The component along the line of sight becomes:
vPS(t) = (1/2) Γ (a v0 / β(aΒ² sinΒ²(Οt) + LΒ²)) sin(2Οt)
Thus, statement (A) is correct.
The observed frequency is given by the Doppler effect, which depends on the relative velocity component vPS(t).
Hence, statement (B) is correct.
The correct statements are (A) and (B).
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 Γβ°(t) and average amplitude A(t) of the system change with time t. Which one of the following options schematically depicts these changes correctly?
At a particular temperature T, Planck's energy density of black body radiation in terms of frequency is \(\rho_T(\nu) = 8 \times 10^{-18} \text{ J/m}^3 \text{ Hz}^{-1}\) at \(\nu = 3 \times 10^{14}\) Hz. Then Planck's energy density \(\rho_T(\lambda)\) at the corresponding wavelength (\(\lambda\)) has the value \rule{1cm}{0.15mm} \(\times 10^2 \text{ J/m}^4\). (in integer)
[Speed of light \(c = 3 \times 10^8\) m/s]
(Note: The unit for \(\rho_T(\nu)\) in the original problem was given as J/mΒ³, which is dimensionally incorrect for a spectral density. The correct unit J/(mΒ³Β·Hz) or JΒ·s/mΒ³ is used here for the solution.)