Differential Equations

Second Order Equations

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Mechanical Oscillations

Solved Problems

Example 1.

A ring of radius \(R\) performs small oscillations around the pivot point \(O\) (Figure \(6\)). Determine the period of oscillation.

Solution.

A ring performing small oscillations
Figure 6.

The ring, suspended at the point \(O,\) is a physical pendulum. The period of oscillation is determined by the formula

\[T = 2\pi \sqrt {\frac{I}{{mga}}} ,\]

where \(I\) is the moment of inertia of the ring about its center, \(m\) is the mass of the ring, \(a\) is the distance from the pivot point to the centre of the ring.

The moment of inertia of a ring of mass \(m\) is equal to \({I_0} = m{R^2}.\) As the distance from the centre of the ring to the pivot point is equal to \(R,\) then using the parallel axis theorem (aka Huygens-Steiner theorem), we find the total moment of inertia of the pendulum:

\[I = {I_0} + m{R^2} = m{R^2} + m{R^2} = 2m{R^2}.\]

Given that \(a = R,\) we obtain the following expression for the oscillation period:

\[T = 2\pi \sqrt {\frac{I}{{mga}}} = 2\pi \sqrt {\frac{{2\cancel{m}{R^{\cancel{2}}}}}{{\cancel{m}g\cancel{R}}}} = 2\pi \sqrt {\frac{{2R}}{g}} .\]

Example 2.

A mass is suspended on two springs connected in series. The stiffness of one spring is twice more than of the other: \({k_2} = 2{k_1}.\) How does the period of oscillation change if the springs are connected in parallel (Figure \(7\))?

Solution.

A mass is suspended on two springs connected in series and in parallel
Figure 7.

We calculate the equivalent stiffness in the case of serial and parallel connection of springs.

In the case of series connection, the elastic force in each spring is equal to the force of gravity (without taking into account the weight of the springs). The total extension is the sum of the extensions of each spring:

\[x = {x_1} + {x_2}.\]

Then the equivalent stiffness is given by

\[x = {x_1} + {x_2},\;\; \Rightarrow \frac{F}{k} = \frac{F}{{{k_1}}} + \frac{F}{{{k_2}}},\;\; \Rightarrow \frac{1}{k} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}},\;\; \Rightarrow k = \frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}.\]

When connected in parallel, the extension of both springs is the same, and the total elastic force will be equal to the sum of the forces in each spring:

\[x = {x_1} = {x_2},\;\;\; F = {F_1} + {F_2}.\]

Hence, the equivalent stiffness of the springs connected in parallel is given by

\[F = {F_1} + {F_2},\;\; \Rightarrow kx = {k_1}{x_1} + {k_2}{x_2} = \left( {{k_1} + {k_2}} \right)x,\;\; \Rightarrow k = {k_1} + {k_2}.\]

The period of oscillation of the springs connected in series is

\[{T_1} = 2\pi \sqrt {\frac{m}{k}} = 2\pi \sqrt {\frac{{m\left( {{k_1} + {k_2}} \right)}}{{{k_1}{k_2}}}} ,\]

and in the case of parallel connection:

\[{T_2} = 2\pi \sqrt {\frac{m}{k}} = 2\pi \sqrt {\frac{m}{{{k_1} + {k_2}}}} .\]

Now we can find how the period of oscillation changes when transitioning from series to parallel connection of the springs:

\[\frac{{{T_2}}}{{{T_1}}} = \frac{{\sqrt {\frac{1}{{{k_1} + {k_2}}}} }}{{\sqrt {\frac{{{k_1} + {k_2}}}{{{k_1}{k_2}}}} }} = \frac{{\sqrt {{k_1}{k_2}} }}{{{k_1} + {k_2}}}.\]

Given that the stiffness of one spring is twice more than the other, we obtain:

\[\frac{{{T_2}}}{{{T_1}}} = \frac{{\sqrt {{k_1}{k_2}} }}{{{k_1} + {k_2}}} = \frac{{\sqrt {2k_1^2} }}{{{k_1} + 2{k_1}}} = \frac{{\sqrt 2 \cancel{k_1}}}{{3\cancel{k_1}}} = \frac{{\sqrt 2 }}{3}.\]

Example 3.

Find the \(Q\) factor of an oscillator, if after \(50\) oscillations the amplitude of the displacement has decreased twice.

Solution.

We first calculate the logarithmic decrement \(\delta.\) By definition, the logarithmic decrement is proportional to the natural logarithm of the ratio of the amplitudes \({x_0}\) and \({x_N}\) of two oscillations, separated by \(N\) periods:

\[\delta = \frac{1}{N}\ln \frac{{{x_0}}}{{{x_N}}}.\]

In our case it is equal to

\[\delta = \frac{1}{{50}}\ln 2 \approx \frac{1}{{50}} \cdot 0,693 = 0,0139.\]

Then the Q factor of the system is

\[Q = \frac{\pi }{\delta } \approx \frac{\pi }{{0,0139}} \approx 227.\]
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