Question

A particle exeuting S.H.M. of amplitude 6cm has an acceleration 16cm/s2 when it is 4cm

from the mean position. Find its period, maximum velocity and maximum acceleration​

Answer 1

⭐《ANSWER》

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↪Actually welcome to the concept of the SHM

↪Basically here , the partical is following the SHM , then the conditions are

↗For Acceleration, a = -w^2 X

↗For time period = 2pi/w

↗For Max! Velocity = Aw

↗For Max! ACCELRATION = A w^2

↪"w" here refers to angular velocity

↗Since here ,

↪the general equation of harmonic motion is

↪X = A sin (wt + ⊙ )

↪Since here ,

↗X = displacement from mean position

↗A = Amplitude of the particle

↗w = Angular frequency

↗⊙ = initial phase / epoch

↗t = time

Answer 2

displacement x= A sin( wt+a)

velocity = dx/dt= Aw cos(wt+a)

acceleration= dv/dt=-Aw^2 sin( wt+a)

= - w^2 x

acceleration magnitude= 16 cm/s^2 when it is 4 cm from mean position

w^2 x= 16

w^2 (4)= 16

w^2 = 4

Max acceleration = w^2 A ( Amplitude is maximum displacement)

= 4× 6 = 24 cm/s^2

Max velocity = Aw

as velocity = Awcos( wt+ a)

Max value of cos( wt+a) = 1

So Aw = 6( 2) = 12 cm/s

Period = 2 pie/w = 2 pie/2 = pie seconds

⭐✌Thanks⭐✌