Two springs have the same modulus of elasticity, 30 N, but are of natural lengths 0.4 m and 0.6 m. An object of mass 0.5 kg is attached to one end of each spring and the other ends are attached to two points which are 1.2 m apart on a smooth horizontal table. Find the period of small oscillations

Question

Two springs have the same modulus of elasticity, 30 N, but are of natural lengths 0.4 m and 0.6 m. An object of mass 0.5 kg is attached to one end of each spring and the other ends are attached to two points which are 1.2 m apart on a smooth horizontal table. Find the period of small oscillations of this system.

Accepted Answer

The vertical forces acting on the object, its weight and the normal reaction of the table, have no effect on the motion: they balance each other.
The first step is to find the equilibrium position of the object. The diagram shows the relevant lengths and the horizontal forces acting when the object is in equilibrium.
The extensions in the springs are [latex]e_{1}[/latex] m and [latex]e_{2}[/latex] m and the tension, T N is the same on both sides.

Hooke’s law applied to each spring gives

T = [latex]\frac{30}{0.4} × e_{1} = 75e_{1}[/latex]

and T = [latex]\frac{30}{0.6} × e_{2} = 50e_{2}[/latex]

⇒ [latex]75e_{1} = 50e_{2}[/latex]

⇒ [latex]e_{2} = 1.5e_{1}[/latex].

The total length is 1.2 m, so [latex]e_{1} + e_{2} = 0.2,[/latex]

⇒ [latex]2.5e_{1} = 0.2.[/latex].

Hence [latex]e_{1} = 0.08 ext{ and } e_{2} = 0.12.[/latex]

The next step is to find the equation of motion, so it is necessary to consider a general position for the object, and this is given by the displacement x m from the equilibrium position. The direction towards the right is taken to be positive.

The tensions in the springs are now different.

Hooke’s law gives

[latex]T_{1} = \frac{30}{0.4} (0.08 + x) = 75(0.08 + x)[/latex]

[latex]T_{2} = \frac{30}{0.6} (0.12 - x) = 50(0.12 - x)[/latex]

Newton’s second law can now be applied giving

[latex]T_{2} - T_{1} = m\ddot{x}[/latex]

⇒ 50(0.12 - x) - 75(0.08 + x) = 0.5[latex]\ddot{x}[/latex]

⇒ -125x = 0.5[latex]\ddot{x}[/latex]

⇒ [latex]\ddot{x}[/latex] = -250x.

This is the equation of motion for SHM with [latex]ω = \sqrt{250}. [/latex]

The period is [latex]\frac{2π}{ω}[/latex] or 0.40 s (to 2 d.p.).

Historical note
The story of the quest for a means to measure longitude shows why scientists at the time of Hooke were interested in both astronomy and accurate time keeping. The problem was solved by John Harrison whose increasingly accurate clocks are preserved at the old Greenwich Observatory. You can find out more about Hooke and Harrison at the website of the National Maritime Museum: www.nmm.ac.uk.

Question 14.9: Two springs have the same modulus of elasticity, 30 N, but a...

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