Question

The average speed over the period of a complete oscillation of a particle performing rectilinear simpleharmonic motion is 2 cm/s. If the particle attains this speed when it is at a point P whose distance from thecentre O is 4 cm, determine(a) The amplitude(b) The periodic time of the motion expressing the answers in terms of t.(c) Show also that the least time taken by the particle to travel from O to P is given bysin-/1​

Answer 1

(a) Amplitude $\boxed{A=\frac{4\pi}{\sqrt{\pi^2-4}}}$

(b) The periodic time of the motion

$\boxed{T=\frac{8\pi}{\sqrt{\pi^2-4}}}$

Explanation:

General equation of SHM

$y=A\sin\omega t$

Where A is the amplitude

Average speed = total distance/Total time

$\implies 2=4A/T$

$\implies T=2A$ ............................. (a)

$\implies \frac{2\pi}{\omega}=2A$

$\implies \omega = \frac{\pi}{A}$

At y = 4 cm, velocity is 2 cm/s

velocity is given by

$v=\frac{dy}{dt}$

$\implies v=A\omega\cos\omega t$

At time t

If y=4 cm then

$4=A\sin \omega t$    ........... (1)

Also at time t

If v=2 cm/s, then

$2=A\omega\cos\omega t$

$\implies \frac{2}{\omega}=A\cos\omega t$  .......(2)

From (1) and (2)

$4^2+(\frac{2}{\omega})^2=A^2(\sin^2\omega t+\cos^2\omega t)$

$\implies 4^2+\frac{2^2A^2}{\pi^2}=A^2$

$\implies A^2-\frac{4A^2}{\pi^2}=4^2$

$\implies A^2(1-\frac{4}{\pi^2})=4^2$

$\implies A^2(\frac{\pi^2-4}{\pi^2}=4^2$

$\implies A^2=\frac{4^2\pi^2}{\pi^2-4}$

$\implies \boxed{A=\frac{4\pi}{\sqrt{\pi^2-4}}}$

(b) From eq (a), Time period

$T=2A$

$\implies \boxed{T=\frac{8\pi}{\sqrt{\pi^2-4}}}$

(c) From eq (1)

$4=A\sin \omega t$

$\implies 4=\frac{4\pi}{\sqrt{\pi^2-4}}\sin \omega t$

$\implies \sin \omega t=\frac{\sqrt{\pi^2-4}}{\pi}$

$\implies \omega t=\sin^{-1}(\frac{\sqrt{\pi^2-4}}{\pi})$

$\implies \frac{2\pi}{T} \times t=\sin^{-1}(\frac{\sqrt{\pi^2-4}}{\pi})$

$\implies t=\frac{T}{2\pi}\sin^{-1}(\frac{\sqrt{\pi^2-4}}{\pi})$

$\implies \boxed{t=\frac{4}{\sqrt{\pi^2-4}}\sin^{-1}(\frac{\sqrt{\pi^2-4}}{\pi})}$    

                                                                                        (Proved)

Hope this answer is helpful.

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