Question

A particle executes SHM with amplitude 0.5 cm and frequently 100sec inverse. the maximum speed of the particle is​

Answer 1

Answer:

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Answer 2

Answer:

A particle executes SHM with amplitude 0.5 cm and frequency 100sec inverse, the maximum speed of the particle is​ $0.031cm/s$

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 $T=2\pi /w$

• The frequency of oscillation is $f=1/T$

From the question, we have

amplitude of oscillation  $A=0.5cm$

frequency $f=1/100sec$

the time period of oscillation

$T=2\pi /w=100s$

⇒angular frequency $w=2\pi /T=2\pi /100=0.062rad/sec$

the maximum speed $V_{max}=Aw=0.5*0.062=0.031cm/s$