Question

$\huge{\tt{Ello!!!}}$
❄️A particle executes SHM on a straight line path. The amplitude of oscillation is 2 cm. When the displacement of the particle from the mean postion is 1 cm,the magnitude of its acceleration is equal to that of velocity. Find the time period of SHM, also the max. velocity and max. acceleration of SHM.
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Answer 1

$\Huge{\boxed{\mathcal{\black{yo}}}}$

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Let y is the displacement y = a sin wt

w = 2π/T

Displacement y = 1

Amplitude a = 2

so 1 = 3 sin 2πt/T

acceleration in SHM is a = w²y

velocity in SHM is v = w√(a²-y²)...1

v = w

4π²/T² = 2π/T (√a² - 1

2π/T = (√3)

T = 2π/√3

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Answer 2

Ello!!!

Here's your answer, check out the attachment.

Thanks!