Q.A particle performing linear S.H.M. of period $2\pi$ seconds about the mean position O is observed to have a speed of $b\sqrt3$ m/s, when at a distance b (metre) from O. If the particle is moving away from O at that instant, find the time required by the particle, to travel a further distance b.

[!TLDR] $\omega=2\pi/T=1$ rad/s. From $v=\omega\sqrt{A^2-x^2}$ at $x=b$, $v=b\sqrt3$: $A=2b$. Time from $x=b$ to $x=A=2b$ is $\Delta t=\frac1\omega\left[\sin^{- …

Given period $T=2\pi$ s, so $\omega=\frac{2\pi}{T}=\frac{2\pi}{2\pi}=1$ rad/s. At $x=b$, speed is $v=b\sqrt3$; using $v=\omega\sqrt{A^2-x^2}$ with $\omega=1$: $b\sqrt3=\sqrt{A^2-b^2}\Rightarrow3b^2=A^2-b^2\Rightarrow A^2=4b^2\Rightarrow A=2b$. So the particle, currently at $x=b$ and moving AWAY from O, must travel a further distance b to reach $x=2b=A$ -- i.e. exactly to the extreme position. Using $x=A\sin(\omega t+\phi_0)$ measured from whatever instant x=b corresponds to, the time to go from displacement $x_1=b=A/2$ to $x_2=2b=A$ is $$\Delta t=\f …

First find A from the given (x,v) pair using $v=\omega\sqrt{A^2-x^2}$, then use the difference of $\sin^{-1}(x/A)$ values (divided by …

Assuming the 'further distance b' takes the particle only to x=2b without checking that this is exactly the extreme position (x=A), …