The maximum displacement of an oscillating particle is 0.05m. If its time period is 1.57s
(i) What is the velocity at the mean position ?
(ii) What is its acceleration at the extreme position ? 

Given,

Time period T = 1.57sec

we know that angular frequency $\mathrm{\omega }=\frac{2\mathrm{\pi }}{\mathrm{T}}$

$\mathrm{\omega }=\frac{2\times 3.14}{1.57}$

$\omega =4$

maximum displacement = n = 0.05m

(i) Velocity will be maximum at the mean position.

Hence velocity at the mean position

${\mathrm{v}}_{\max }=A\omega$

${\mathrm{v}}_{\max }=0.05\times 4$

${\mathrm{v}}_{\max }=0.2\mathrm{m}/\mathrm{s}$

velocity at mean position will be 0.2m/s.

(ii) Acceleration at extreme position a

$\mathrm{a}={\mathrm{\omega }}^2\mathrm{A}$

$\mathrm{a}=4^2\mathrm{A}$

$\mathrm{a}=16\times 0.05$

$\mathrm{a}=0.8\mathrm{m}/{\mathrm{s}}^2$

Hence the correct answer is $0.2 {\mathrm{ms}}^{-1}, 0.8{\mathrm{ms}}^{-2}.$