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 =  \(\frac{1}{2}v_max\) , 2A=10 cm

 A=5 cm

v = \(ω\sqrt{A^2-x^2}\)   and v_max = ωA

Since v = \(\frac{1}{2}v_max\)

\(ω\sqrt{A^2-x^2}=\frac{ωA}{2}\)

∴ \(A^2-x^2=\frac{A^2}{4}\)

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

∴ \(x= ±\frac{\sqrt{3}}{2}A\)

= 0.866 x 5= ± 4.33 cm

This gives the required displacement.