Question

A particle of mass 1kg is moving with SHM. Its greatest velocity is 20m/s and it's amplitude is 10m.find the period and force of attraction towards the center when the particle is at its greatest distance

Answer 1

The mass of the particle [m] = 1 kg.

The amplitude of SHM  [r] = 10 m.

The maximum velocity [v] =20 m/s

We know that maximum velocity  $v=r\omega$

                                                        $\omega=\frac{v}{r}$

                                                        $\omega=\frac{20}{10} \ rad/s$

                                                                            = 2 rad/s

The maximum acceleration of a simple harmonic motion    $[a]=-\omega^2r$

The condition for simple harmonic motion is that acceleration is just opposite to the displacement.That's why the negative sign is taken.

                                                                             $=-2^{2} *10\ m/s^2$

                                                                             $=-40\ m/s^2$

Now we have to calculate the time period. The time period is calculated as-

                                           $T=\frac{2\pi}{\omega}$

                                              $=\frac{2\pi}{2}$

                                              $=\ \pi\ seconds$

Now we are asked to calculate the force when the particle is at greatest distance i.e that is at the extreme point.

The force F = ma

                   = 1 kg×[-40] N

                   = -40 N

Here negative sign indicates that the force is opposite to the displacement.

The magnitude of the force will be 40 N.