12.4. Oscillations - 1

A particle is executing S.H.M. of amplitude 5 cm and period of 2 sec. Find the speed of particle at a point where its acceleration is half of its maximum value.

Given-

Amplitude             =       A       =       5 cm  = 5 X 10^-2 m

Time period           =       T       =       2 sec

                                        

Simple Harmonic Motion-

              Simple harmonic motion is a type of periodic motion or oscillation motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.  

The acceleration of particle performing SHM is given ( numerically ) by

 a = ω² x

i.e.

x = a / ω²    

_ _ _ _ _ _ _ _ _ _ ( 1 )

Here we are considering a point where the acceleration is half of maximum value

i.e.

a = a _max / 2

 _ _ _ _ _ _ _ _ _ _ ( 2 )

w. k. t. at extreme position acceleration is maximum and it can be given as

a _max= ± A ω²    

 _ _ _ _ _ _ _ _ _ _ ( 3 )

from equation ( 2 ) and ( 3 ) , equation ( 1 ) becomes

x = A ω² / 2 ω²

x = A / 2

The speed of particle performing SHM is

V = ω √ ( A² – x² )

As,

ω  = 2 π / T

Where T is a time period

Putting values of ω and x we get

V = [ ( 2 π ) / T ] √ [ A² – ( A / 2 ) ² ]

V = [ ( 2 π ) / T ] √ [ A² – ( A²  / 4 ) ]

V = [ ( 2 π ) / T ] √ { A² [ 1 – ( 1 / 4 ) ] }

V = [ ( 2 π ) / T ] √ ( 3 / 4  ) A²

 V = [ ( 2 π ) / 2 T ] A √ ( 3 )

 V = [ ( π ) / T ] A √ ( 3 )

Putting the values of A and T we get,

V = [ ( π ) / 2 ] ( 5 X 10^-2  ) √ ( 3 )

V = 13.6 X 10^-2  m / s

V = 13.6 cm / s

Therefore the speed of particle at a point where its acceleration is half of its maximum value is V = 13.6 cm / s