Question

if the maximum acceleration of a particle executing SHM is 2m and it's amplitude is 1/2m then angular frequency is​

Answer 1

Answer:

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Answer 2

Answer:

If the maximum acceleration of a particle executing SHM is $2m/s^{2}$ and its amplitude is 1/2m then the angular frequency is​ $2rad/sec$

Explanation:

• In simple harmonic motion, the displacement of a particle is given by the formula

                  $x(t)=Asin(\omega t+\phi )$

         where

        $A$-amplitude

       ω-angular frequency

       Ф-initial phase

• The velocity of that particle is

                 $dx(t)/dt=Awsin(\omega t+\phi )$  

• The acceleration of that particle is

                 $d^{2} x(t)/dt^{2} =Aw^{2} sin(\omega t+\phi )$

         maximum acceleration is

                 $a_{max} =Aw^{2}$

From the question, we have

the amplitude $A=1/2m$

the maximum acceleration $a_{max} =Aw^{2}=2m/s^{2}$

substitute the value of the amplitude, we get

$1/2w^{2} =2\\w^{2}=4\\w=2 rad /sec$

the angular frequency ω=$2 rad/sec$