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).
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 Hookeβs 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):