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Q.

A particle executes simple harmonic motion with an amplitude of 5cm. When the particle is at 4cm from the mean position. The magnitude of its velocity in SI units is equal to that of its acceleration. Then its periodic time in seconds is

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a

$\frac{4 π}{3}$

b

$\frac{3 π}{8}$

c

$\frac{8 π}{3}$

d

$\frac{7 π}{3}$

answer is A.

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Detailed Solution

$A m p l i t u d e A = 5 cm (given) Velocity of particle = acceleration of particle at displacement y = 4 \Rightarrow ω \sqrt{A^{2} - y^{2}} = ω^{2} y, h e r e w i s a n g u l a r v e l o c i t y \Rightarrow \sqrt{5^{2} - 4^{2}} = ω .4 \frac{3}{4} = ω = \frac{2 π}{T} \Rightarrow t i m e p e r i o d i s T = \frac{8 π}{3} s$

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