Question

a particle execute shm along a straight line . the amplitude of oscillation is 2cm. when displacement of particle from the mean position is 1cm the magnitude of its acceleration is equal to magnitude of its velocity . the time period of oscillation is?

Answer 1

As we know that
A = Amplitude = 2 cm
x =displacement = 1 cm
a = acceleration =-w^2x
v =velocity = w (A^2-x^2)^1/2
as Magnitude of a = Magnitude of b i.e
w^2x = w (A^2-x^2)^1/2
implies
wx = (A^2-x^2)^1/2
also 2 = 2pi/T
therefore
2pi/T = (A^2-x^2)1/2
T = 2pi/(A^2-x^2)1/2
T = 2pi/(4-1)^1/2
T = 2pi/sqrt(3)
T = 1.73 seconds
Check if this answer is correct , keep me posted :)