Question

The graph below shows the motion of an unforced undamped harmonic oscillator:

Answer 1

The simple harmonic motion and Hooke's law allows to find the displacement when hanging the mass of 2 kg is:

      x = 2.45 m

Simple harmonic motion is a motion where the restoring force is proportional to the displacement.

          x = A cos (wt + Ф)

Where x is the displacement, A is the amplitude, w is the angular velocity, t is the time

The angular velocity is

          w² = k / m

Where k is the spring constant and m is the mass.

From the graph we see that the oscillatory wave has half a period between the times t₁ = $\frac{3}{4} \pi$  and t₂ = $\frac{1}{4} \pi$ , therefore the period is:

                T = π  s.

The angle in a complete oscillation is θ = 2π rad.

The angular kinematics defines the angular velocity with the angle in the unit of time.

        w = $\frac{\Delta \theta}{\Delta t}$  

        w = $\frac{2\pi }{\pi }$  

        w = 2 rad / s

Let's find the spring constant.

        k = w² m

Let's calculate.

        k = 2² 2

        k = 8 N / m

Hooke's law says that the restoring force of a spring is proportional to its displacement.

         F  = - k x

         mg = - k x

          x = $\frac{mg}{k}$  

          x = $\frac{2 \ 9.8}{8}$  

          x = 2.45 m

In conclusion using simple harmonic motion and Hooke's law we can find the displacement when hanging the 2kg mass is:

      x = 2.45 m

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