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 ? - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
0.2 ms^–1, 0.2m^s–2
B
$0.8 {ms}^{- 1}, 0.8 {ms}^{- 2}$
C
$0.2 {ms}^{- 1}, 0.8 {ms}^{- 2}$
D
$0.8 {ms}^{- 1}, 0.2 {ms}^{- 2}$

Solution:

Max. displacements = Amplitude
A = 0.05m ; T = 1.57sec
→velocity at mean position
$\begin{array}{l} v_{max} = ωA = \frac{2 π}{T} \times A = \frac{2 π}{T} \times A = 2 \times \frac{3 .14}{1 .57} \times 0 .05 \\ = 4 \times 0 .05 = 0 .2 m / \sec \\ \to Acceleration at extreme position \\ a = - ω^{2} A [- ve sign only direction] \\ a = {[\frac{2 π}{T}]}^{2} \times A = {[\frac{2 \times 3 .14}{1 .57}]}^{2} \times 0 .05 \\ = 16 \times 0 .05 = 0 .8 m / \sec^{2} \end{array}$

Solution:

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