Question

Please help!!
A particle executes shm with time period 2s and amplitude 5cm. What is the maximum magnitude of velocity??

Answer 1

I think 5 cm/ sec is the maximum velocity of the particle

Answer 2

Answer:

A particle executes SHM with time period 2s and amplitude 5cm, the maximum magnitude of the velocity is $15.7cm/s$

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  $A=5cm$

the time period of oscillation

$T=2\pi /w=2s$

⇒angular frequency $w=2\pi /T=2\pi /2=3.14rad/sec$

the maximum speed $V_{max}=Aw=5*3.14=15.7cm/s$