Question

The figure given below depicts two circular motions. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated in the figures. Obtain the simple harmonic motions of the x-projection of the radius vector of the rotating particle P in each case.​

Answer 1

✨$\huge\mathcal\color{Green} AnSwEr::$

(a) At t = 0, OP makes an angle of 45°= π/4 with the (positive direction of) x-axis. After

$time \: t \: it \: covers \: an \: angle \: \frac{2\pi}{T}t \: in \: the \: \\ anticlockwise \: sense \: and \: makes \: an \\ \: angle \: of \: \frac{2\pi}{T}t \: + \frac{\pi}{4} \: with \: the \ \\ x - axis.$

The projection of OP on the x-axis at time t is given by,

$x(t) = </em><em>A</em><em>\cos ( \frac{2\pi}{</em><em>T</em><em>}t + \frac{\pi}{4})$

For T = 4s,

$x(t) = </em><em>A</em><em>\cos ( \frac{2\pi}{</em><em>T</em><em>}t + \frac{\pi}{4})$

Which is a SHM of amplitude A, period 4 s, and an initial phase* = π/4.

(b) In this case at t= 0, OP makes an angle of 90°= π/2 with the x-axis. After a time t, it covers an angle of $\frac{2\pi}{T}t$ in the clockwise sense and makes an angle of $( \frac{\pi}{2} - \frac{2\pi}{T}t)$

In this case at t= 0, OP makes an angle of 90°= π/2 with the x-axis. After a time t, it covers an angle of $\frac{2\pi}{T}t$ in the clockwise sense and makes an angle of $( \frac{\pi}{2} - \frac{2\pi}{T}t)$with the x-axis.The projection of OP on the x-axis at time t is given by....

$x(t) = B \cos( \frac{\pi}{2} - \frac{2\pi}{T}t )$

$= B \sin( \frac{2\pi}{T}t)$

For T = 30 s,

$x(t) = B \sin( \frac{\pi}{15}t)$

$writing \: this \: as \: x(t) = B \cos( \frac{\pi}{15}t - \frac{\pi}{2})$

We find that this represents a SHM of

amplitude B, period 30 s, and and initial phase of -π/2.

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