Q.At what distance from the mean position is the speed of a particle performing S.H.M. half its maximum speed? Given path length of S.H.M. = 10 cm.

[!TLDR] Path length 10 cm means amplitude A = 5 cm. At v = v_max/2, x = (sqrt(3)/ …

The 'path length' of an S.H.M. is the total distance from one extreme to the other, i.e. 2A, NOT the amplitude itself. Given path length = 10 cm, the amplitude is $A=\frac{10}{2}=5$ cm. Using $v=\omega\sqrt{A^2-x^2}$, at the mean position $v_{max}=\omega A$; we want the displacement where $v=v_{max}/2$. Setting $\omega\sqrt{A^2-x^2}=\frac{\omega A}{2}$ …

Recognise path length = 2A to first find A = 5 cm, then apply $x=\frac{\sqrt3}{2}A$ (fro …

Using the given path length (10 cm) directly as the amplitude instead of halving it firs …