Question

A particle of mass 1 kg, executing SHM has amplitude 10 cm and time penod 2 s. The maximum speed of the particle is (in m/s)​

Answer 1

Answer:

ANSWER

v

max

=Aw

=20cm× 2π

T

=0.2m× 2π

1s

=1.256m/s

I hope it's help you:)))

Answer 2

Answer:

A particle of mass 1 kg, executing SHM has an amplitude of 10 cm and time period 2 s. The maximum speed of the particle is $31.4cm/s$

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  $A=10cm$

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 $V_{max}=Aw=10*3.14=31.4cm/s$