a particle executes linear shmbwith an amplitude if 2 cm.when the particle is at 1cm from the mean position the magnitude of its velocity is equal to that of its acceleration.Then its time period is...??

velocity at distance x in shm is given as 

v= w sqrt(A2-x2)

so A = 2 x =1

v = w sqrt(4-1) = wsqrt3

now acc at distance x is a= w2 x = w2

as mag a = mag v

w2 = wsqrt3

w= sqrt3

2pi/T= sqrt3

T = 2pi/sqrt3

V = W (A² - X² )^1/2 

A=W²X

WHERE ''W'' IS ANGULAR FREQUENCY LET ''T'' BE TIME PERIOD T=2PIE/W

X=0.01m AND A=0.02m

ACCORDING TO THE GIVEN CONDITION 

V=A 

THERE FORE 

W (A² - X² )^1/2 =W²X

FIND THE VALUE OF ''W'' AND GET YOUR ANSWER 

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