A particle of mass 200 g is attached to a light spring of natural length 40 cm and stiffness 50 $Nm^{-1}$ . The particle is allowed to hang vertically in equilibrium.
i) Find the extension of the spring in this position.
The spring is now pulled down 3 cm and released from rest.
ii) Find the length of the spring as a function of time.

Question

A particle of mass 200 g is attached to a light spring of natural length 40 cm and stiffness 50 [latex]Nm^{-1}[/latex]. The particle is allowed to hang vertically in equilibrium.
i) Find the extension of the spring in this position.
The spring is now pulled down 3 cm and released from rest.
ii) Find the length of the spring as a function of time.

Accepted Answer

i) The diagram shows the relevant lengths, and the forces acting on the particle in equilibrium. The extension in the spring is [latex]x_{0}[/latex] m.

By Hooke’s law: T = k[latex]x_{0}[/latex] = 50[latex]x_{0}[/latex].

Since the particle is in equilibrium,

T = 0.2g

⇒ 50[latex]x_{0}[/latex] = 0.2g

[latex]x_{0}[/latex] = 0.0392.

The extension is 0.0392 m = 3.92 cm.

ii) Let x m be the displacement from the equilibrium position in the downwards direction.

[latex]\begin{matrix} ext{The extension of the spring is then } x + x_{0} & \boxed{ ext{You know the value of }x_{0}.}\end{matrix}[/latex]

By Hooke’s law, [latex]T = 50(x + x_{0}).[/latex]

Applying Newton’s second law in the downwards direction,

0.2g - 50(x + [latex]x_{0}[/latex] ) = 0.2[latex]\ddot{x}[/latex]

[latex]\begin{matrix}⇒ 0.2g - 50x - 50x_{0} = 0.2\ddot{x}. & \longleftarrow \boxed{ ext{This shows the advantage of measuring x from the equilibrium position. The weight and }kx_{0} ext{ cancel each other out}} \end{matrix}[/latex]

However, as found in part i), 0.2g = 50[latex]x_{0}[/latex] and so the equation may be simplified to

[latex]\begin{matrix}0.2\ddot{x} = - 50x & \longleftarrow \boxed{ ext{This shows the advantage of measuring x from the equilibrium position. The weight and }kx_{0} ext{ cancel each other out}} \end{matrix}[/latex]

⇒ [latex]\ddot{x} = - 250x.[/latex]

This is the SHM equation with ω² = 250, and so [latex]ω = \sqrt{250}.[/latex]

Since the particle starts at the end of the oscillation with x = 0.03, the appropriate equation is of the form x = a cos ωt, with a = 0.03 and [latex]ω = \sqrt{250}.[/latex]

So, at time t, x = 0.03 cos [latex]\sqrt{250}t[/latex]

and the total length is 0.4 + 0.0392 + 0.03 cos [latex]\sqrt{250}t[/latex]

= 0.4392 + 0.03 cos [latex]\sqrt{250}t[/latex].

Question 14.8: A particle of mass 200 g is attached to a light spring of na...

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