Question

A particle executing SHM has amplitude 3cm and maximum velocity 6.24cm/s, what is the time period of the particle?

Answer 1

Explanation:

please see the attached picture dear

Answer 2

Answer:

A particle executing SHM has an amplitude of 3cm and a maximum velocity of 6.24cm/s, the time period of the particle is $3.01s$

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  $A=3cm$

the maximum speed $V_{max}=Aw=6.24cm/s$

⇒angular frequency $w=v_{max} /A=6.24/3=2.08rad/sec$

the time period of oscillation

$T=2\pi /w=\frac{2*3.14}{2.08}=3.01s$