Question

What is the maximum speed of a particle executing SHM WITH amplitude of 3cm and periodic time 6second.​

Answer 1

Refer to the attachment

Hope it helps you

Answer 2

Answer:

The maximum speed of a particle executing SHM with an amplitude of 3cm and period of time is 6s is $3.14*10^{-2} m/s$

Explanation:

• In simple harmonic motion restoring force is directly proportional to the acceleration of the body

• The displacement of the body executing SHM is given by the formula

                       $x(t)=A sin(wt+$Ф)

         where,

        $A$-amplitude of the oscillation

        ω-the angular frequency

        Ф-phase

• The speed of the particle is $v(t)=dx/dt=A wsin(wt+$Ф), and the maximum speed is when the sine value is one $V_{max}=Aw$

• The time period is the time taken to complete one oscillation, it is given by the formula $T=2\pi /w$

From the question, we have

amplitude of oscillation  $A=3cm=0.03m$

the time period of oscillation $T=6s$

⇒

$T=2\pi /w=6\\w=2\pi /6\\w=\pi /3$

⇒the maximum speed $V_{max}=Aw=0.03*\pi /3=3.14*10^{-2}m/s$