Step 1. \(v=\omega\sqrt{a^2\,-\,y^2},\)
\(v_2=\omega\sqrt{a^2\,-\,y_1^2}\)
And \(v_2 =\omega\sqrt{a^2\,-y^2_2}\)
4 = \(\omega\sqrt{a^2\,-\,3^2}\) ...(i)
And 3 = \(\omega\sqrt{a^2\,-\,4^2}\) ...(ii)
Dividing eqn (i) by (ii) :
\(\frac{4}{3}=\frac{\sqrt{a^2-9}}{\sqrt{a^2-16}}\)
i.e., \(\frac{16}{9}=\frac{\sqrt{a^2-9}}{\sqrt{a^2-16}}\)
Or 16a2 − 256 = 9a2 − 81
Or a2= 25
Or a = 5 m.
Step 2. Putting a = 5 in eqn (i),
4 = 4ω
Or ω = 1 rad s-1
Step 3. Time taken to travel half amplitude from positive extreme positive extreme position is given by displacement,
x =acos ωt
i.e., \(\frac{5}{2}= 5cos(1\times t) \,or\,cos\, t=\frac{1}{2}\)
Or t = cos-1\(\frac{1}{2}\) = 60º = \(\frac{\pi}{3}\)
Or t = \(\frac{3.142}{3}\) = 1.047 s.