Question

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.

Answer 1

Given : v =  $\frac{1}{2}{v}_{m}ax$ , 2A=10 cm

 A=5 cm

v = $\omega \sqrt{A^2-x^2}$   and v_max = ωA

Since v = $\frac{1}{2}{v}_{m}ax$

$\omega \sqrt{A^2-x^2}=\frac{\omega A}{2}$

∴ $A^2-x^2=\frac{A^2}{4}$

∴ $x^2=A^2-\frac{A^2}{4}=\frac{3A^2}{4}$

∴ $x=\pm \frac{\sqrt{3}}{2}A$

= 0.866 x 5= ± 4.33 cm

This gives the required displacement.