OSCILLATIONS-Notes
Physics - Notes
Periodic Motion
A motion is called periodic if a system repeats its motion after equal intervals of time. The smallest interval of time after which the motion repeats itself is called the time period of the motion.
Mathematically,
if a physical quantity describing motion has the same value at time tand \(t+T\), where Tis constant, the motion is periodic.
\[x(t)=x(t+T)\]Important Characteristics of Periodic Motion
- The system returns to its initial state after every fixed time interval.
- The time period remains constant for ideal periodic motion.
- The path followed may or may not be along a straight line.
- Not all periodic motions involve to-and-fro movement about a fixed position.
Oscillatory Motion
Oscillatory motion is a special type of periodic motion in which a body moves to and fro about a fixed position, called the mean position, under the influence of a restoring tendency.
Thus, every oscillatory motion is periodic, but every periodic motion is not oscillatory.
Essential Features of Oscillatory Motion
- The motion takes place about a stable equilibrium (mean) position.
- The body moves alternately on both sides of the mean position.
- A restoring force always acts on the body, directed towards the mean position.
- The motion repeats itself after equal intervals of time.
Mean Position
The mean position is the position where the net force acting on the body is zero. At this position:
- Acceleration of the particle is zero.
- Potential energy is minimum.
- Velocity is maximum during oscillation.
Extreme Positions
The extreme positions are the farthest positions on either side of the mean position. At these points:
- Velocity becomes zero.
- Acceleration is maximum.
- The restoring force is maximum in magnitude.
Difference Between Periodic and Oscillatory Motion
| Periodic Motion | Oscillatory Motion |
|---|---|
| Repeats after equal intervals of time | To-and-fro motion about a mean position |
| May not involve restoring force | Always involves restoring force |
| May not pass through equilibrium | Always passes through mean position |
| Example: Circular motion | Example: Pendulum motion |
Period and frequency
Period (Time Period)
The period or time period of an oscillatory motion is defined as the time taken by a particle to complete one full oscillation, that is, to return to its initial position with the same direction of motion.
It is generally denoted by the symbol \(T\).
Frequency
The frequency of an oscillatory motion is defined as the number of complete oscillations performed per second.
It is usually denoted by the symbol \(f\) or \(\nu\).
Mathematical Relation Between Period and Frequency
If a particle completes one oscillation in time \(T\), then it completes
\[\frac{1}{T}\]Oscillations in one second.
By definition, frequency is the number of oscillations per second. Hence,
\[f=\frac{1}{T}\]or equivalently,
\[\boxed{\;T=\frac{1}{f}\;}\]Displacement
Displacement in oscillatory motion is defined as the directed distance of the oscillating particle from its mean position at any given instant of time, measured along the line of motion.
It is usually denoted by the symbol \(x\).
Nature of Displacement
Displacement is a vector quantity, as it possesses both magnitude and direction.
- When the particle is on one side of the mean position, displacement is taken as positive.
- When it is on the opposite side, displacement is taken as negative.
The choice of direction is a matter of convention, but once chosen, it must be followed consistently.
Basic Periodic Displacement Function
One of the simplest and most important periodic functions used to describe oscillatory motion is the cosine function:
\[f(t)=A\cos\,\omega t\]
where:
\(A\) is the maximum value of displacement (amplitude),
\(\omega\) is the angular frequency,
\(t\) is time.
Period of the Cosine Function
The cosine function repeats its value whenever its argument increases by an integral multiple of \(2\pi\).
Thus, the condition for repetition is:
\[\omega(t+T)=\omega t+2\pi\]Cancelling \omega \(t\), we get:
\[\omega T=2\pi\]Hence, the time period is:
\[T=\frac{2\pi}{\omega}\]Therefore, the displacement function satisfies:
\[f(t)=f(t+T)\]Sine Function as a Periodic Displacement
The same reasoning applies to a sine function:
\[f(t)=A\sin\,\omega t\]This function is also periodic with the same time period:
\[T=\frac{2\pi}{\omega}\]Linear Combination of Sine and Cosine Functions
A more general form of periodic displacement is:
\[f(t)=A\sin\,\omega t+B\cos\,\omega t\]This function is also periodic with time period:
\[T=\frac{2\pi}{\omega}\]Conversion into Single Sine Function
By choosing:
\[\begin{aligned}A&=D\cos\,\phi\quad\text{and,}\\\\B&=D\sin\,\phi\end{aligned}\]the above expression can be written as:
\[f(t)=D\sin\,(\omega t+\phi)\]where:
\[\begin{aligned}D&=\sqrt{A^2+B^2},\\\\\tan\,\phi&=\frac{B}{A}\end{aligned}\]
Here:
\(D\) is the resultant amplitude,
\(\phi\) is the phase constant.
Simple Harmonic Motion
A particle is said to execute simple harmonic motion if its acceleration is directly proportional to its displacement from the mean position and is always directed towards the mean position.
Mathematically,
\[a\propto-x\] or, \[a=-\omega^2x\]
where:
\(x\) is the displacement from the mean position,
\(a\) is the acceleration,
\(\omega\) is a positive constant called the angular frequency,
the negative sign indicates that acceleration is opposite to displacement.
Physical Conditions for SHM
For a motion to be simple harmonic, the following conditions must be satisfied:
- The system must have a stable equilibrium position.
- When displaced, a restoring force must act on the particle.
- The restoring force must be directly proportional to the displacement.
- The restoring force must always be directed towards the mean position.
If any of these conditions is not fulfilled, the motion will not be SHM.
Derivation of the Equation of SHM (Force Approach)
Consider a particle of mass mdisplaced by a small distance xfrom its mean position. Suppose a restoring force acts on it such that:
\[F\propto-x\]Introducing a proportionality constant \(k\),
\[F=-kx\]Using Newton’s second law,
\[ \begin{aligned} F&=ma\\ ma&=-kx\\ a&=-\frac{k}{m}x \end{aligned} \]Comparing with the standard SHM form \(a=-\omega^2x\), we get:
\[\omega^2=\frac{k}{m}\]Thus, the acceleration is proportional to displacement and directed towards the mean position, proving that the motion is simple harmonic.
SHM as a Periodic Motion (Proof)
Statement
Every simple harmonic motion is periodic.
Proof
In SHM, the displacement of the particle varies with time as:
\[x=A\sin\,(\omega t+\phi)\] or \[x=A\cos\,(\omega t+\phi)\]Sine and cosine functions repeat their values after a fixed interval of time \(T\). Since the displacement returns to the same value after each interval \(T\), the motion repeats itself. Hence, simple harmonic motion is periodic.
Time Period of SHM
The argument of sine or cosine increases by 2\piin one complete oscillation. Thus,
\[\begin{aligned} \omega T&=2\pi\\\\ T&=\frac{2\pi}{\omega} \end{aligned} \]The time period of SHM depends only on the angular frequency and is independent of displacement and time.
Displacement, Velocity, and Acceleration in SHM
Displacement
\[x=A\sin\,(\omega t+\phi)\]Velocity
Velocity is the rate of change of displacement:
\[v=\frac{dx}{dt}=A\omega\cos\,(\omega t+\phi)\]Velocity is maximum at the mean position and zero at the extreme positions.
Acceleration
Acceleration is the rate of change of velocity:
\[a=\frac{dv}{dt}=-A\omega^2\sin\,(\omega t+\phi)\]Thus,
\[a=-\omega^2x\]This confirms that acceleration is proportional to displacement and opposite in direction, which is the defining condition of SHM.
Simple Harmonic Motion and Uniform Circular Motion
Simple Harmonic Motion
A particle executes simple harmonic motion if its acceleration is:
- directly proportional to its displacement from the mean position, and
- always directed towards the mean position.
Uniform Circular Motion
A particle is said to execute uniform circular motion if it moves along a circular path with constant speed.
Although the speed remains constant, the velocity continuously changes direction. Therefore, the particle experiences an acceleration directed towards the centre of the circle, called centripetal acceleration.
\[a_c=\frac{v^2}{r}=\omega^2r\]
where:
\(r\) is the radius of the circle,
\(\omega\) is the angular speed.
Displacement of the Projection
Let \(M\) be the projection of point \(P\) on the diameter along the x-axis.
The displacement of \(M\) from the centre \(O\) is:
\[x=A\cos\,\theta\]Substituting \(\theta=\omega t\),
\[x=A\cos\,\omega t\]This is the standard equation of displacement in SHM.
Velocity of the Projected Motion
The velocity of the projected particle Mis given by:
\[ \begin{aligned} v&=\frac{dx}{dt}\\\\ v&=-A\omega\sin\,\omega t \end{aligned} \]
- Velocity is maximum at the mean position.
- Velocity is zero at the extreme positions.
Acceleration of the Projected Motion
Differentiating velocity with respect to time:
\[a=\frac{dv}{dt}=-A\omega^2\cos\,\omega t\]Since \(x=A\cos\,\omega t\),
\[a=-\omega^2x\]This proves that:
- acceleration is proportional to displacement,
- acceleration is always directed towards the mean position.
VELOCITY AND ACCELERATION IN SIMPLE HARMONIC MOTION
Velocity in Simple Harmonic Motion
The velocity of a particle executing SHM is defined as the rate of change of displacement with respect to time.
\[v(t)=\frac{dx(t)}{dt}\]Derivation Using Reference Circle Method
To obtain a clear physical picture, SHM can be visualised as the projection of a uniform circular motion on a diameter.
Consider a particle moving uniformly in a circle of radius \(A\) with angular speed \(\omega\). The speed of the particle in circular motion is:
\[v=\omega A\]The direction of this velocity is always tangential to the circle.
Let the projection of this particle on a diameter represent a particle executing SHM. At time \(t\), if the angular position is \(\omega t +\phi\), then from geometry:
\[v(t)=-\omega A\sin\,(\omega t+\phi)\]The negative sign indicates that the direction of velocity may be opposite to the chosen positive direction of the axis.
This expression gives the instantaneous velocity of a particle executing SHM.
Analytical Derivation of Velocity
If the displacement in SHM is given by:
\[x(t)=A\cos\,(\omega t+\phi)\]Differentiating with respect to time:
\[v(t)=\frac{dx}{dt}=-\omega A\sin\,(\omega t+\phi)\]Thus, velocity varies sinusoidally with time.
Derivation Using Reference Circle Method
In uniform circular motion, the particle experiences centripetal acceleration directed towards the centre:
\[a_c=\frac{v^2}{A}=\omega^2A\]This acceleration acts along the radius towards the centre. The component of this acceleration along the diameter gives the acceleration of the projected particle executing SHM.
Thus,
\[a(t)=-\omega^2A\cos\,(\omega t+\phi)\]Since displacement is:
\[x(t)=A\cos\,(\omega t+\phi)\]we get:
\[a(t)=-\omega^2x(t)\]Special Case: Taking Phase Constant \(\phi=0\)
- Displacement:
\[x(t)=A\cos\,\omega t\] - Velocity:
\[v(t)=-\omega A\sin\,\omega t\] - Acceleration:
\[a(t)=-\omega^2A\cos\,\omega t\]These equations describe the complete kinematics of SHM.
For simplicity, if the motion starts from the extreme position, we write:
FORCE LAW FOR SIMPLE HARMONIC MOTION
Using Newton’s second law of motion, and the expression for acceleration of a particle undergoing SHM, the force acting on a particle of mass \(m\) in SHM is
\[ \begin{aligned} F (t) &= ma\\ &= –m\omega^2 \times (t)\\\\ F (t) &= –k x (t) \end{aligned} \]where \(k = m\omega^2\)
or \[\boxed{\;\omega=\sqrt{\dfrac{k}{m}}\;}\]ENERGY IN SIMPLE HARMONIC MOTION
Both kinetic and potential energies of a particle in SHM vary between zero and their maximum values.
It is zero at the extreme positions of displacement. Therefore, the kinetic energy (K) of such a particle, which is defined as
\[ \begin{aligned} K&=\dfrac{1}{2}mv^2\\\\ &=\dfrac{1}{2}m\omega^2A^2\sin^2\,(\omega t+\phi)\\\\ &=\dfrac{1}{2}kA^2\sin^2\,(\omega t+\phi) \end{aligned} \]is also a periodic function of time, being zero when the displacement is maximum and maximum when the particle is at the mean position. Note, since the sign of v is immaterial in K, the period of K is T/2.
potential energy is possible only for conservative forces. The spring force \(F = –kx\) is a conservative force, with associated potential energy
\[U=\dfrac{1}{2}kx^2\]Hence the potential energy of a particle executing simple harmonic motion is,
\[ \begin{aligned} U(x)&=\dfrac{k}{x^2}\\\\ &=\dfrac{1}{2}kA^2\cos^2(\omega t + \phi) \end{aligned} \]Thus, the potential energy of a particle executing simple harmonic motion is also periodic, with period T/2, being zero at the mean position and maximum at the extreme displacements.
The total energy, E, of the system is,
\[ \begin{aligned} E&=U+K\\\\ &=\dfrac{1}{2}kA^2\cos^2(\omega t + \phi)+\dfrac{1}{2}kA^2\sin^2(\omega t + \phi)\\\\ &=\dfrac{1}{2}kA^2\Bigl[cos^2(\omega t + \phi)+sin^2(\omega t + \phi)\Bigr] \end{aligned} \]Using the familiar trigonometric identity, the value of the expression in the brackets is unity. Thus,
\[\boxed{\;E=\dfrac{1}{2}kA^2\;}\]The Simple Pendulum
Let \(\theta) be the angle made by the string with the vertical. When the bob is at the mean position, \(\theta = 0\)
There are only two forces acting on the bob; the tension \(T\) along the string and the vertical force due to gravity (=mg).
The force mg can be resolved into the component mg cosθ along the string and mg sinθ perpendicular to it.
Since the motion of the bob is along a circle of length L and centre at the support point, the bob has a radial acceleration \((\omega^2 L)\) and also a tangental acceleration; the latter arises since motion along the arc of the circle is not uniform.
The radial acceleration is provided by the net radial force \(T –mg \cos \theta\), while the tangential acceleration is provided by \(mg \sin \theta\)
Torque \(\tau\) about the support is entirely provided by the tangental component of force
\[\tau=-L(mg \sin \theta)\]This is the restoring torque that tends to reduce angular displacement — hence the negative sign. By Newton’s law of rotational motion,
\[\tau=I\alpha\]where I is the moment of inertia of the system about the support and α is the angular acceleration. Thus,
\[I \alpha = –m g \sin \theta L\] or \[\alpha=-\dfrac{mgL}{I}\sin \theta\]if we assume that the displacement \(\theta\) is small. We know that \(\sin \theta\) can be expressed as,
\[\sin \theta=\theta -\dfrac{\theta^3}{3!} + \dfrac{\theta^5}{5!}\pm\cdots\]Now if \(\theta\) is small, \(\sin \theta\) can be approximated by \(\theta\)
\[\alpha=-\dfrac{mgL}{I}\theta\] \[\omega=\sqrt{\dfrac{mgL}{I}}\] \[T=2\pi\sqrt{\dfrac{I}{mgL}}\]Example-1
On an average, a human heart is found to beat 75 times in a minute. Calculate its frequency and period.
Solution
The average rate of beating of the human heart is given as 75 beats per minute. Frequency is defined as the number of events occurring per second, so the given rate must first be converted from minutes to seconds.
Using the definition of frequency,
\[ \begin{aligned} \text{Frequency},\; f &= \frac{\text{Number of beats per minute}}{\text{Number of seconds in one minute}} \\ &= \frac{75}{60} \\ &= 1.25\ \text{Hz} \end{aligned} \]
Thus, the frequency of the heartbeat is \(1.25\ \text{Hz}\), meaning the heart beats 1.25 times every second.
The time period is defined as the time taken for one complete cycle, and it is the reciprocal of frequency. Therefore,
\[ \begin{aligned} \text{Time period},\; T &= \frac{1}{f} \\ &= \frac{1}{1.25} \\ &= 0.8\ \text{s} \end{aligned} \]
Hence, the frequency of the human heartbeat is \(1.25\ \text{Hz}\) and the corresponding time period is \(0.8\ \text{s}\).
Example-2
Which of the following functions of time represent
(a) periodic and
(b) non-periodic motion? Give the period for each case of periodic motion [ω is any positive constant].
(i) sin ωt + cos ωt
(ii) sin ωt + cos 2 ωt + sin 4 ωt
(iii) e–ωt
(iv) log (ωt)
Solution
A motion is said to be periodic if it repeats itself after a fixed interval of time called the time period. Each given function is examined on this basis.
For the function \( \sin \omega t + \cos \omega t \), both sine and cosine terms have the same angular frequency \( \omega \). Since each term individually repeats after a time period \( \frac{2\pi}{\omega} \), their sum also repeats after the same interval. Hence, this motion is periodic with time period
\[ \begin{aligned} T &= \frac{2\pi}{\omega} \end{aligned} \]
For the function \( \sin \omega t + \cos 2\omega t + \sin 4\omega t \), the angular frequencies involved are \( \omega \), \( 2\omega \), and \( 4\omega \). The corresponding time periods are \( \frac{2\pi}{\omega} \), \( \frac{\pi}{\omega} \), and \( \frac{\pi}{2\omega} \). Since these periods are integral multiples of the smallest period \( \frac{\pi}{2\omega} \), the function repeats itself after
\[ \begin{aligned} T &= \frac{2\pi}{\omega} \end{aligned} \]
Therefore, this motion is also periodic with time period \( \frac{2\pi}{\omega} \).
For the function \( e^{-\omega t} \), the value continuously decreases with time and never repeats its earlier values. Since there is no fixed time interval after which the motion repeats, this motion is non-periodic.
Similarly, for the function \( \log(\omega t) \), the value keeps changing monotonically with time and does not repeat itself for any finite interval. Hence, this motion is also non-periodic.
In summary, the functions in cases (i) and (ii) represent periodic motion with time period \( \frac{2\pi}{\omega} \), while the functions in cases (iii) and (iv) represent non-periodic motion.
Example-3
Which of the following functions of time represent
(a) simple harmonic motion and
(b) periodic but not simple harmonic? Give the period for each case.
(1) sin ωt – cos ωt
(2) sin2 ωt
Solution
A motion is said to be simple harmonic if it can be expressed as a single sine or cosine function of time with a constant angular frequency. If a motion is periodic but cannot be reduced to this standard form, it is classified as periodic but not simple harmonic.
For the function \( \sin \omega t - \cos \omega t \), both sine and cosine terms have the same angular frequency \( \omega \). This expression can be combined into a single sinusoidal form using trigonometric identities. Writing it in amplitude–phase form,
\[ \begin{aligned} \sin \omega t - \cos \omega t &= \sqrt{2}\,\sin\left(\omega t - \frac{\pi}{4}\right) \end{aligned} \]
Since the displacement varies sinusoidally with time and involves only one angular frequency, this motion represents simple harmonic motion. The time period is therefore given by
\[ \begin{aligned} T &= \frac{2\pi}{\omega} \end{aligned} \]
For the function \( \sin^{2}\omega t \), the displacement is not a simple sine or cosine function of time. Using the identity \( \sin^{2}\theta = \frac{1 - \cos 2\theta}{2} \), the expression becomes
\[ \begin{aligned} \sin^{2}\omega t &= \frac{1 - \cos 2\omega t}{2} \end{aligned} \]
This shows that the motion is periodic, but it does not satisfy the condition of simple harmonic motion because it is not directly proportional to a single sine or cosine of \( \omega t \). The angular frequency involved is \( 2\omega \), and hence the time period is
\[ \begin{aligned} T &= \frac{2\pi}{2\omega} = \frac{\pi}{\omega} \end{aligned} \]
Thus, \( \sin \omega t - \cos \omega t \) represents simple harmonic motion with period \( \frac{2\pi}{\omega} \), while \( \sin^{2}\omega t \) represents periodic but not simple harmonic motion with period \( \frac{\pi}{\omega} \).
Example-4
A body oscillates with SHM according to the equation (in SI units), x = 5 cos [2π t + π/4].
At t = 1.5 s, calculate the
(a) displacement,
(b) speed and (
c) acceleration of the body.
Solution
The given equation of motion of the body executing simple harmonic motion is \[ x = 5\cos\left(2\pi t + \frac{\pi}{4}\right), \] where the displacement \(x\) is in metres and time \(t\) is in seconds. Comparing this equation with the standard SHM form \(x = A\cos(\omega t + \phi)\), the amplitude is \(A = 5\ \text{m}\) and the angular frequency is \(\omega = 2\pi\ \text{rad s}^{-1}\).
At \(t = 1.5\ \text{s}\), the phase angle becomes
\[ \begin{aligned} \omega t + \phi &= 2\pi(1.5) + \frac{\pi}{4} \\ &= 3\pi + \frac{\pi}{4} \\ &= \frac{13\pi}{4}. \end{aligned} \]
The displacement at this instant is therefore
\[ \begin{aligned} x &= 5\cos\left(\frac{13\pi}{4}\right) \\ &= 5\cos\left(\frac{5\pi}{4}\right) \\ &= 5\left(-\frac{1}{\sqrt{2}}\right) \\ &= -\frac{5}{\sqrt{2}}\ \text{m} \approx -3.54\ \text{m}. \end{aligned} \]
The velocity in SHM is given by \(v = \frac{dx}{dt} = -A\omega \sin(\omega t + \phi)\). Substituting the given values,
\[ \begin{aligned} v &= -5(2\pi)\sin\left(\frac{13\pi}{4}\right) \\ &= -10\pi \sin\left(\frac{5\pi}{4}\right) \\ &= -10\pi\left(-\frac{1}{\sqrt{2}}\right) \\ &= \frac{10\pi}{\sqrt{2}}\ \text{m s}^{-1} \approx 22.2\ \text{m s}^{-1}. \end{aligned} \]
The acceleration in SHM is given by \(a = -\omega^{2}x\). Using the value of displacement already calculated,
\[ \begin{aligned} a &= -(2\pi)^{2}\left(-\frac{5}{\sqrt{2}}\right) \\ &= \frac{20\pi^{2}}{\sqrt{2}}\ \text{m s}^{-2} \approx 139.6\ \text{m s}^{-2}. \end{aligned} \]
Hence, at \(t = 1.5\ \text{s}\), the displacement of the body is approximately \(-3.54\ \text{m}\), the speed is about \(22.2\ \text{m s}^{-1}\), and the acceleration is approximately \(1.40 \times 10^{2}\ \text{m s}^{-2}\).
Example-5
A block whose mass is 1kg is fastened to a spring. The spring has a spring constant of 50 N m–1. The block is pulled to a distance x = 10 cm from its equilibrium position at x = 0 on a frictionless surface from rest at t = 0. Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position.
Solution
A block of mass \(m = 1\ \text{kg}\) is attached to a spring of spring constant \(k = 50\ \text{N m}^{-1}\). The block is displaced from its equilibrium position by \(10\ \text{cm}\) and released from rest. This maximum displacement represents the amplitude of oscillation, so \(A = 0.10\ \text{m}\).
In simple harmonic motion of a spring–block system, the total mechanical energy remains constant and is equal to the maximum potential energy stored in the spring at the extreme position. Hence, the total energy of the system is
\[ \begin{aligned} E &= \frac{1}{2}kA^{2} \\ &= \frac{1}{2}\times 50 \times (0.10)^{2} \\ &= 25 \times 0.01 \\ &= 0.25\ \text{J}. \end{aligned} \]
When the block is \(5\ \text{cm}\) away from the mean position, its displacement is \(x = 0.05\ \text{m}\). The potential energy stored in the spring at this position is
\[ \begin{aligned} U &= \frac{1}{2}kx^{2} \\ &= \frac{1}{2}\times 50 \times (0.05)^{2} \\ &= 25 \times 0.0025 \\ &= 0.0625\ \text{J}. \end{aligned} \]
Since the total mechanical energy is conserved, the kinetic energy of the block at this position is obtained by subtracting the potential energy from the total energy,
\[ \begin{aligned} K &= E - U \\ &= 0.25 - 0.0625 \\ &= 0.1875\ \text{J}. \end{aligned} \]
Therefore, when the block is \(5\ \text{cm}\) away from the mean position, its potential energy is \(0.0625\ \text{J}\), its kinetic energy is \(0.1875\ \text{J}\), and the total mechanical energy of the system remains constant at \(0.25\ \text{J}\).
Example-6
What is the length of a simple pendulum, which ticks seconds ?
Solution
A seconds pendulum is defined as a simple pendulum that takes one second to move from one extreme position to the other and another second to return, so its time period is \(T = 2\ \text{s}\).
For a simple pendulum undergoing small oscillations, the time period is given by
\[ \begin{aligned} T &= 2\pi \sqrt{\frac{l}{g}} \end{aligned} \]
Rearranging this expression to find the length \(l\),
\[ \begin{aligned} l &= \frac{gT^{2}}{4\pi^{2}} \end{aligned} \]
Substituting \(T = 2\ \text{s}\) and taking the acceleration due to gravity as \(g = 9.8\ \text{m s}^{-2}\),
\[ \begin{aligned} l &= \frac{9.8 \times (2)^{2}}{4\pi^{2}} \\ &= \frac{39.2}{4\pi^{2}} \\ &\approx 0.994\ \text{m}. \end{aligned} \]
Hence, the length of a simple pendulum that ticks seconds is approximately \(1.0\ \text{m}\).
