Academia Aeternum
तमसो मा ज्योतिर्गमय
Crystal-clear concepts, exam-oriented derivations, and formula shortcuts for mastering periodic motion — from atomic vibrations to the ticking of a pendulum.
Oscillations represent one of the most fundamental motions in physics — a repetitive back-and-forth movement about an equilibrium position. From the swing of a playground pendulum to vibrations in atoms and molecules, oscillatory motion governs countless natural phenomena.
At the heart of this chapter lies Simple Harmonic Motion (SHM) — a special class of oscillation where the restoring force is directly proportional to displacement. This elegant relationship, F = −kx, unlocks the mathematics of waves, sound, and even quantum states.
A strong grasp of oscillations builds the foundation for advanced topics like waves, alternating current, and quantum mechanics — making this one of the most strategically important chapters in the entire curriculum.
Understand motion that repeats after equal intervals of time — the definition of period, frequency, and displacement functions.
Learn the conditions for SHM and derive displacement, velocity, and acceleration equations from first principles.
Study kinetic energy, potential energy, and conservation of total mechanical energy across the full oscillation cycle.
Analyze oscillations produced by elastic restoring forces — including series and parallel spring combinations.
Understand how gravitational restoring force produces periodic oscillations and how period depends on length, not mass.
Explore real-world oscillations where energy gradually dissipates due to resistive forces like air drag and friction.
Oscillations is among the most frequently tested chapters in IIT-JEE, NEET, and all major engineering and medical entrance examinations. Its problems blend conceptual depth with numerical precision — a combination that separates top rankers from the rest.
Start by understanding the restoring force concept. Once you grasp why F = −kx creates oscillation, every formula becomes logical rather than something to memorize.
Practice displacement–time, velocity–time, and acceleration–time graphs. Many entrance exam questions specifically test graphical reading rather than direct formula substitution.
Focus on the fundamental SHM equation set above. Once mastered thoroughly, almost all numerical problems reduce to quick substitutions.
Advanced questions combine oscillations with energy conservation, spring systems, or inclined planes. Solve previous year papers to develop speed and pattern recognition.
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.
Let \(x(t)\) be a physical quantity (such as position, angle, or displacement) describing the motion. If this quantity has the same value at times \(t\) and \(t + T\), where \(T\) is a constant, then the motion is periodic.
\[ x(t) = x(t + T) \]Oscillatory motion is a special type of periodic motion in which a body moves to and fro about a fixed position, called the mean or equilibrium position, under the action of a restoring tendency.
In such motion, whenever the body is displaced from its mean position, a restoring force or torque acts on it, always directed towards the mean position.
Thus, every oscillatory motion is periodic, but every periodic motion is not oscillatory (for example, uniform circular motion is periodic but not oscillatory).
For small displacements in many oscillatory systems, the restoring force is approximately proportional to the displacement and opposite in direction, which can be written as \(F \propto -x\). Such motion is often called simple harmonic motion.
The mean position (or equilibrium position) in oscillatory motion is the position of the particle where the net force acting on it is zero. In this position, the particle would remain at rest if it were placed there and not given any push.
During an oscillation, the particle passes through the mean position every cycle while moving from one extreme position to the other.
The extreme positions (also called turning points) in oscillatory motion are the farthest positions of the particle on either side of the mean (equilibrium) position. At these points, the displacement from the mean position is maximum.
The particle momentarily comes to rest at each extreme position, then reverses its direction of motion and moves back towards the mean position.
Oscillatory motion is a special type of periodic motion. All oscillatory motions are periodic, but there are many periodic motions that are not oscillatory (for example, uniform circular motion).
| Periodic Motion | Oscillatory Motion |
|---|---|
| Any motion that repeats itself after equal intervals of time. | A periodic motion in which the body moves to and fro about a fixed mean (equilibrium) position. |
| May involve linear, circular, or any curved path. | Path is along a line or a fixed arc about the mean position (to‑and‑fro motion). |
| May or may not involve a restoring force. | Always involves a restoring force (or torque) directed towards the mean position. |
| The motion may not pass through a single fixed equilibrium point (e.g., a particle in uniform circular motion). | The body necessarily passes through the same mean position every cycle. |
| Examples: motion of the Earth around the Sun, hands of a clock, uniform circular motion of a fan blade. | Examples: motion of a simple pendulum (for small angles), mass–spring system, vibrations of a tuning fork. |
| It is the broader category of repeating motions. | It is a sub‑category of periodic motion with a to‑and‑fro character and a mean position. |
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 \(T\).
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 \(\nu\).
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,
\[\boxed{\;T=\frac{1}{f}\;}\]In oscillatory motion, displacement is defined as the directed distance of the oscillating particle from its mean (equilibrium) position at any instant of time, measured along the line of motion.
It is usually denoted by the symbol \(x\).
Displacement is a vector quantity, as it has both magnitude and direction along the line of motion.
The choice of which side is positive is a matter of convention, but once chosen, it must be followed consistently.
A very important model for oscillatory motion is a sinusoidal displacement. A simple form 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.
The cosine function repeats its value whenever its argument increases by an integral multiple of \(2\pi\).
For \(f(t) = A \cos(\omega t)\), 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 the periodicity condition:
\[ f(t) = f(t + T) \]Similarly, a sine function can describe displacement:
\[ f(t) = A \sin(\omega t) \]This function is also periodic with the same time period:
\[ T = \frac{2\pi}{\omega} \]A more general periodic displacement with the same angular frequency can be written as:
\[ f(t) = A \sin(\omega t) + B \cos(\omega t) \]This function is also periodic with time period:
\[ T = \frac{2\pi}{\omega} \]By choosing:
\[ \begin{aligned} A &= D \cos\phi, \\ B &= D \sin\phi \end{aligned} \]the above expression can be rewritten as a single sinusoidal function:
\[ f(t) = D \sin(\omega t + \phi) \]where:
\[ \begin{aligned} D &= \sqrt{A^2 + B^2}, \\ \tan\phi &= \frac{B}{A} \end{aligned} \]Here \(D\) is the resultant amplitude and \(\phi\) is the phase constant (or initial phase) of the motion.
A particle is said to execute simple harmonic motion (SHM) if its acceleration is directly proportional to its displacement from the mean (equilibrium) position and is always directed towards the mean position.
In SHM, the acceleration \(a\) of the particle is proportional to its displacement \(x\) from the mean position and opposite in direction:
\[ a \propto -x \]or, introducing the constant \(\omega^2\),
\[ a = -\omega^2 x \]
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 shows that acceleration is opposite to displacement.
For a motion to be simple harmonic, the following conditions must be satisfied:
If any of these conditions is not fulfilled, the motion will not be SHM.
Consider a particle of mass \(m\) displaced by a small distance \(x\) from 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 \(a = -\omega^2 x\), 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.
Every simple harmonic motion is periodic.
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 time interval \(T\). Since the displacement returns to the same value after each interval \(T\), the motion repeats itself. Hence, simple harmonic motion is periodic.
The argument of sine or cosine increases by \(2\pi\) in one complete oscillation. Thus:
\[ \begin{aligned} \omega T &= 2\pi \\ T &= \frac{2\pi}{\omega} \end{aligned} \]The time period \(T\) of SHM depends only on the angular frequency \(\omega\) and is independent of displacement and time.
A standard expression for displacement in SHM is:
\[ x = A \sin(\omega t + \phi) \]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 is the rate of change of velocity:
\[ a = \frac{dv}{dt} = -A \omega^2 \sin(\omega t + \phi) \]Thus,
\[ a = -\omega^2 x \]This confirms that acceleration is proportional to displacement and opposite in direction, which is the defining condition of SHM.
A particle executes simple harmonic motion (SHM) if its acceleration is:
Mathematically,
\[ a = -\omega^2 x \]A particle is said to execute uniform circular motion (UCM) if it moves along a circular path with constant speed.
Although the speed remains constant, the velocity continuously changes its direction. Therefore, the particle experiences an acceleration directed towards the centre of the circle, called centripetal acceleration.
For a particle moving in a circle of radius \(r\) with constant speed \(v\) and angular speed \(\omega\), the centripetal acceleration is:
\[ a_c = \frac{v^2}{r} = \omega^2 r \]
where:
\(r\) is the radius of the circle,
\(\omega\) is the angular speed.
Consider a particle \(P\) moving with uniform angular speed \(\omega\) in a circle of radius \(A\). Let \(M\) be the projection of \(P\) on a diameter taken as the \(x\)-axis, and \(O\) the centre of the circle.
If the angular position of \(P\) is \(\theta\), then the displacement of \(M\) from the centre \(O\) is:
\[ x = A \cos \theta \]For uniform circular motion, \(\theta = \omega t\). Hence,
\[ x = A \cos(\omega t) \]This is the standard equation of displacement in SHM.
The velocity of the projected particle \(M\) is:
\[ \begin{aligned} v &= \frac{dx}{dt} \\ &= \frac{d}{dt}\big[A \cos(\omega t)\big] \\ &= -A \omega \sin(\omega t) \end{aligned} \]From this expression:
These are characteristic features of SHM.
Differentiating velocity with respect to time:
\[ a = \frac{dv}{dt} = \frac{d}{dt}\big[-A \omega \sin(\omega t)\big] = -A \omega^2 \cos(\omega t) \]Since \(x = A \cos(\omega t)\), we can write:
\[ a = -\omega^2 x \]This shows that:
Hence, the motion of the projection of a particle in uniform circular motion on any diameter is simple harmonic motion.
The velocity of a particle executing SHM is defined as the rate of change of displacement with respect to time:
\[ v(t) = \frac{dx(t)}{dt} \]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 the velocity of the projection is:
\[ v(t) = -\omega A \sin(\omega t + \phi) \]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.
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 and changes sign as the particle moves from one side of the mean position to the other.
In uniform circular motion, the particle experiences a centripetal acceleration directed towards the centre:
\[ a_c = \frac{v^2}{A} = \omega^2 A \]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.
Hence, for the projection:
\[ a(t) = -\omega^2 A \cos(\omega t + \phi) \]Since the displacement is:
\[ x(t) = A \cos(\omega t + \phi) \]the acceleration can also be obtained by differentiating velocity:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}\big[-\omega A \sin(\omega t + \phi)\big] = -\omega^2 A \cos(\omega t + \phi) \]Therefore,
\[ a(t) = -\omega^2 x(t) \]This relation shows that acceleration is proportional to displacement and always directed towards the mean position, which is the defining condition of SHM.
For simplicity, if the motion starts from an extreme position, we can choose \(\phi = 0\). Then:
These equations together describe the complete kinematics (displacement, velocity, and acceleration) of a particle in simple harmonic motion.
In simple harmonic motion, the net force acting on the particle is a restoring force that is directly proportional to the displacement from the mean position and always directed towards the mean position.
Using Newton’s second law and the expression for acceleration in SHM, we can write the force law for a particle of mass \(m\) undergoing SHM.
For SHM, the acceleration of the particle is:
\[ a(t) = -\omega^2 x(t) \]Applying Newton’s second law \(F = ma\) for a particle of mass \(m\):
\[ \begin{aligned} F(t) &= m\,a(t) \\ &= m\big[-\omega^2 x(t)\big] \\ &= -m\omega^2\,x(t) \end{aligned} \]We can now define the constant
\[ k = m\omega^2 \]so that the force law takes the familiar Hooke’s law form:
\[ F(t) = -k\,x(t) \]Here \(k\) is called the force constant or spring constant of the system. The negative sign shows that the restoring force is always directed opposite to the displacement (towards the mean position).
From \(k = m\omega^2\), we obtain the angular frequency of SHM as:
\[ \boxed{\;\omega = \sqrt{\dfrac{k}{m}}\;} \]Thus, for a given mass, a stiffer system (larger \(k\)) oscillates with a larger angular frequency, and hence a smaller time period.
In simple harmonic motion (SHM), both kinetic energy (K) and potential energy (U) vary with time between zero and their respective maximum values, while the total mechanical energy of the system remains constant.
For a particle of mass \(m\) in SHM, the kinetic energy is defined as:
\[ K = \frac{1}{2} m v^2 \]If the displacement is \(x(t) = A \cos(\omega t + \phi)\), then the velocity is \(v(t) = -A\omega \sin(\omega t + \phi)\). Substituting this in the expression for kinetic energy gives:
\[ \begin{aligned} K &= \frac{1}{2} m \big[-A\omega \sin(\omega t + \phi)\big]^2 \\ &= \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t + \phi) \end{aligned} \]Using \(k = m\omega^2\), this can be written as:
\[ K = \frac{1}{2} k A^2 \sin^2(\omega t + \phi) \]Thus, kinetic energy is a periodic function of time, zero at the extreme positions (where \(v = 0\)) and maximum at the mean position (where \(|v|\) is maximum). Since \(K\) depends on \(\sin^2\), its period is \(T/2\).
Potential energy can be defined only for conservative forces. The spring force \(F = -kx\) is conservative and has associated potential energy:
\[ U(x) = \frac{1}{2} k x^2 \]For SHM with \(x(t) = A \cos(\omega t + \phi)\), the potential energy becomes:
\[ \begin{aligned} U(t) &= \frac{1}{2} k \big[x(t)\big]^2 \\ &= \frac{1}{2} k \big[A \cos(\omega t + \phi)\big]^2 \\ &= \frac{1}{2} k A^2 \cos^2(\omega t + \phi) \end{aligned} \]Hence, the potential energy in SHM is also periodic with period \(T/2\), being zero at the mean position (\(x = 0\)) and maximum at the extreme displacements (\(x = \pm A\)).
The total mechanical energy \(E\) of the system is the sum of kinetic and potential energies:
\[ \begin{aligned} E &= U + K \\ &= \frac{1}{2} k A^2 \cos^2(\omega t + \phi) + \frac{1}{2} k A^2 \sin^2(\omega t + \phi) \\ &= \frac{1}{2} k A^2 \big[\cos^2(\omega t + \phi) + \sin^2(\omega t + \phi)\big] \end{aligned} \]Using the identity \(\cos^2\theta + \sin^2\theta = 1\), we obtain:
\[ \boxed{\;E = \dfrac{1}{2} k A^2\;} \]Thus, the total mechanical energy of a simple harmonic oscillator is constant and depends only on the spring constant \(k\) and the amplitude \(A\).
A simple pendulum consists of a small, heavy bob suspended by a light, inextensible string of length \(L\), fixed at its upper end. The bob swings in a vertical plane under gravity.
Let \(\theta\) be the angle made by the string with the vertical. When the bob is at the mean (equilibrium) position, \(\theta = 0\).
There are only two forces acting on the bob: the tension \(T\) along the string and the weight \(mg\) acting vertically downward.
The weight \(mg\) can be resolved into:
The bob moves along a circular arc of radius \(L\) and centre at the support point, so it has:
The radial acceleration is provided by the net radial force \((T - mg \cos\theta)\), while the tangential acceleration is provided by the component \(mg \sin\theta\).
The restoring torque \(\tau\) about the support is entirely due to the tangential component of the weight:
\[ \tau = -L \,(mg \sin\theta) \]The negative sign indicates that this torque tends to reduce the angular displacement (i.e., it is a restoring torque).
By Newton’s law of rotational motion,
\[ \tau = I \alpha \]where \(I\) is the moment of inertia of the system about the support and \(\alpha\) is the angular acceleration. Hence,
\[ I \alpha = -mgL \sin\theta \]or
\[ \alpha = -\dfrac{mgL}{I} \sin\theta \]For small angular displacements, we use the expansion:
\[ \sin\theta = \theta - \dfrac{\theta^3}{3!} + \dfrac{\theta^5}{5!} - \cdots \]When \(\theta\) is small (in radians), higher powers of \(\theta\) are negligible and we can approximate:
\[ \sin\theta \approx \theta \]Then the equation of motion becomes:
\[ \alpha = -\dfrac{mgL}{I}\,\theta \]Comparing with the standard SHM angular form \(\alpha = -\omega^2 \theta\), we identify:
\[ \omega = \sqrt{\dfrac{mgL}{I}} \] and the time period is \[ T = 2\pi \sqrt{\dfrac{I}{mgL}} \]For an ideal simple pendulum with a point mass \(m\) at the end of a massless string of length \(L\), the moment of inertia about the support is:
\[ I = mL^2 \]Substituting in the expression for \(\omega\),
\[ \omega = \sqrt{\dfrac{mgL}{mL^2}} = \sqrt{\dfrac{g}{L}} \]Hence, the time period of small oscillations of a simple pendulum is:
\[ T = 2\pi \sqrt{\dfrac{L}{g}} \]This shows that for small angles, the motion of a simple pendulum is simple harmonic, with period depending only on the length \(L\) and acceleration due to gravity \(g\), and not on the mass of the bob.
Many periodic processes in nature, such as a swinging pendulum or a beating heart, repeat themselves again and again. The frequency of such a process is the number of complete cycles (or events) occurring per second, measured in hertz (Hz). The time period \(T\) is the time taken for one complete cycle. These two quantities are related by \[ f = \frac{1}{T} \quad\text{and}\quad T = \frac{1}{f}. \] Thus, if we know how many times an event occurs in a given time (like “beats per minute”), we can find its frequency and then its period.
On an average, a human heart is found to beat 75 times in a minute. Calculate its frequency and period.
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:
\[ \begin{aligned} \text{Time period},\; T &= \frac{1}{f} \\ &= \frac{1}{1.25} \\ &= 0.8\ \text{s} \end{aligned} \]
Hence, the frequency of the human heartbeat is \(1.25\ \text{Hz}\) and the corresponding time period is \(0.8\ \text{s}\).
In oscillations and waves, a motion (or any quantity describing it) is said to be periodic if it repeats its value after a fixed interval of time called the time period \(T\). Mathematically, a function \(f(t)\) is periodic with period \(T\) if \[ f(t + T) = f(t)\quad\text{for all }t. \] Sine and cosine functions are classic examples: \(\sin(\omega t)\) and \(\cos(\omega t)\) both repeat after a time \(T = \dfrac{2\pi}{\omega}\). Sums of periodic functions can also be periodic, provided their periods are commensurate (integer multiples of a common basic period). Functions that never repeat their values (like exponentials or logarithms) represent non-periodic motion.
Which of the following functions of time represent
(a) periodic and
(b) non-periodic motion? Give the period for each case of periodic motion
[\(\omega\) is any positive constant].
(i) \(\sin \omega t + \cos \omega t\)
(ii) \(\sin \omega t + \cos 2 \omega t + \sin 4 \omega t\)
(iii) \(e^{-\omega t}\)
(iv) \(\log (\omega t)\)
For the function \(\sin \omega t + \cos \omega t\), both sine and cosine terms have the same angular frequency \(\omega\). Each term individually repeats after a time period \(\dfrac{2\pi}{\omega}\), so their sum also repeats after the same interval. Hence, this motion is periodic with time period
\[ T = \frac{2\pi}{\omega}. \]
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 \(\dfrac{2\pi}{\omega}\), \(\dfrac{\pi}{\omega}\), and \(\dfrac{\pi}{2\omega}\), respectively. The smallest of these is \(\dfrac{\pi}{2\omega}\), and the other periods are integer multiples of this smallest period. Therefore, the sum repeats after
\[ T = \frac{2\pi}{\omega}, \]
so this motion is also periodic with time period \(\dfrac{2\pi}{\omega}\).
For the function \(e^{-\omega t}\), the value decreases continuously with time and never returns to any earlier value. There is no finite time interval after which the function repeats identically. Hence, this motion is non-periodic.
For the function \(\log(\omega t)\), the value also changes monotonically with time and does not repeat itself for any finite interval. Therefore, this motion is also non-periodic.
In summary:
A motion is said to be simple harmonic if its displacement can be written in the form \( x(t) = A \sin(\omega t + \phi) \) or \( x(t) = A \cos(\omega t + \phi) \), i.e., a single sine or cosine function with constant amplitude \(A\) and angular frequency \(\omega\). In such a case the acceleration satisfies \(a = -\omega^2 x\). A motion is periodic but not simple harmonic if it repeats after a fixed time \(T\), but the displacement cannot be reduced to a single sinusoidal function of time.
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) \(\sin \omega t - \cos \omega t\)
(2) \(\sin^{2} \omega t\)
For the function \(\sin \omega t - \cos \omega t\), both sine and cosine terms have the same angular frequency \(\omega\). A linear combination of \(\sin\omega t\) and \(\cos\omega t\) with the same frequency can always be written as a single sinusoidal function using amplitude–phase form:
\[ \begin{aligned} \sin \omega t - \cos \omega t &= \sqrt{2}\,\sin\left(\omega t - \frac{\pi}{4}\right) \end{aligned} \]
This is clearly of the standard SHM form \(x(t) = A \sin(\omega t + \phi)\) with amplitude \(A = \sqrt{2}\) and the same angular frequency \(\omega\). Therefore, this function represents simple harmonic motion. The time period is
\[ T = \frac{2\pi}{\omega}. \]
For the function \(\sin^{2}\omega t\), the displacement is not directly a simple sine or cosine function of time. Using the trigonometric identity \(\sin^{2}\theta = \dfrac{1 - \cos 2\theta}{2}\), we can write:
\[ \begin{aligned} \sin^{2}\omega t &= \frac{1 - \cos 2\omega t}{2}. \end{aligned} \]
This shows that the motion is periodic (it is a cosine function plus a constant term), but the dependence on time is through \(\cos 2\omega t\), not a single sinusoidal function of \(\omega t\). The motion does not satisfy the SHM condition \(a = -\omega^2 x\), so it is not simple harmonic.
The angular frequency present is \(2\omega\), so the time period is
\[ T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}. \]
Thus:
In simple harmonic motion, the displacement of a particle can be written in the standard form \[ x(t) = A \cos(\omega t + \phi), \] where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase constant. Once \(x(t)\) is known, the velocity and acceleration at any instant can be obtained by differentiation: \[ v(t) = \frac{dx}{dt} = -A\omega \sin(\omega t + \phi), \quad a(t) = \frac{dv}{dt} = -\omega^2 A \cos(\omega t + \phi) = -\omega^2 x(t). \] Thus, given the SHM equation, we can directly find displacement, velocity, and acceleration at a specified time.
A body oscillates with SHM according to the equation (in SI units)
\(x = 5 \cos \big[2\pi t + \dfrac{\pi}{4}\big]\).
At \(t = 1.5\ \text{s}\), calculate the
(a) displacement,
(b) speed, and
(c) acceleration of the body.
The given equation of motion is \[ x = 5\cos\left(2\pi t + \frac{\pi}{4}\right), \] where \(x\) is in metres and \(t\) in seconds. Comparing with the standard form \(x = A\cos(\omega t + \phi)\), we identify:
At \(t = 1.5\ \text{s}\), the phase angle is
\[ \begin{aligned} \omega t + \phi &= 2\pi(1.5) + \frac{\pi}{4} \\ &= 3\pi + \frac{\pi}{4} \\ &= \frac{13\pi}{4}. \end{aligned} \]
The displacement at this instant is
\[ \begin{aligned} x &= 5\cos\left(\frac{13\pi}{4}\right). \end{aligned} \]
Since \(\dfrac{13\pi}{4} = \dfrac{8\pi}{4} + \dfrac{5\pi}{4} = 2\pi + \dfrac{5\pi}{4}\), and cosine has period \(2\pi\), \[ \cos\left(\frac{13\pi}{4}\right) = \cos\left(\frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}}. \] Therefore, \[ \begin{aligned} x &= 5\left(-\frac{1}{\sqrt{2}}\right) \\ &= -\frac{5}{\sqrt{2}}\ \text{m} \approx -3.54\ \text{m}. \end{aligned} \]
The velocity in SHM is \[ v(t) = \frac{dx}{dt} = -A\omega \sin(\omega t + \phi). \] Substituting the given values,
\[ \begin{aligned} v &= -5(2\pi)\sin\left(\frac{13\pi}{4}\right) \\ &= -10\pi \sin\left(\frac{5\pi}{4}\right) \\ &= -10\pi\left(-\frac{1}{\sqrt{2}}\right) \\ &= \frac{10\pi}{\sqrt{2}}\ \text{m s}^{-1}. \end{aligned} \]
Numerically, \[ v \approx 22.2\ \text{m s}^{-1}. \] (Speed is the magnitude of velocity, so the speed is \(22.2\ \text{m s}^{-1}\).)
The acceleration in SHM can be written as \[ a(t) = -\omega^{2}x(t). \] Using the value of displacement already calculated,
\[ \begin{aligned} a &= -(2\pi)^{2}\left(-\frac{5}{\sqrt{2}}\right) \\ &= \frac{20\pi^{2}}{\sqrt{2}}\ \text{m s}^{-2}. \end{aligned} \]
Numerically, \[ a \approx 1.40 \times 10^{2}\ \text{m s}^{-2}. \]
Hence, at \(t = 1.5\ \text{s}\), the displacement is approximately \(-3.54\ \text{m}\), the speed is about \(22.2\ \text{m s}^{-1}\), and the acceleration is approximately \(1.40 \times 10^{2}\ \text{m s}^{-2}\).
For a block–spring system executing simple harmonic motion, the total mechanical energy is conserved (in the absence of friction). The potential energy is stored in the spring and is given by \(U = \tfrac{1}{2}kx^{2}\), where \(k\) is the spring constant and \(x\) is the displacement from the equilibrium position. The kinetic energy of the mass is \(K = \tfrac{1}{2}mv^{2}\). At any instant, \[ E = K + U = \text{constant} = \frac{1}{2}kA^{2}, \] where \(A\) is the amplitude (maximum displacement). Knowing \(E\) and \(U\) at a given \(x\), we can find \(K = E - U\).
A block whose mass is \(1\ \text{kg}\) is fastened to a spring. The spring has a spring constant of \(50\ \text{N m}^{-1}\). The block is pulled to a distance \(x = 10\ \text{cm}\) 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 \(5\ \text{cm}\) away from the mean position.
Given:
Since the block is pulled to \(0.10\ \text{m}\) and released from rest, this maximum displacement is the amplitude of oscillation: \[ A = 0.10\ \text{m}. \]
For a spring–block SHM, the total energy is equal to the maximum potential energy at the extreme position:
\[ \begin{aligned} E &= \frac{1}{2}kA^{2} \\ &= \frac{1}{2}\times 50 \times (0.10)^{2} \\ &= 25 \times 0.01 \\ &= 0.25\ \text{J}. \end{aligned} \]
When the block is \(5\ \text{cm} = 0.05\ \text{m}\) away from the mean position, the displacement is \(x = 0.05\ \text{m}\). The potential energy stored in the spring 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 at this position is the difference between total and potential 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:
A seconds pendulum (or second's pendulum) is a simple pendulum whose time period is exactly \(2\ \text{s}\): it takes \(1\ \text{s}\) to swing from one extreme to the other and another \(1\ \text{s}\) to return. For small oscillations of a simple pendulum of length \(l\) in a uniform gravitational field of acceleration \(g\), the time period is \[ T = 2\pi \sqrt{\frac{l}{g}}. \] Given the desired period \(T\), this formula can be rearranged to obtain the required length \(l\).
What is the length of a simple pendulum which ticks seconds?
A seconds pendulum is defined as one whose time period is \[ T = 2\ \text{s}. \]
For small oscillations of a simple pendulum, the time period is given by
\[ T = 2\pi \sqrt{\frac{l}{g}}. \]
Rearranging this expression to solve for the length \(l\),
\[ \begin{aligned} T &= 2\pi \sqrt{\frac{l}{g}} \\ \Rightarrow \sqrt{\frac{l}{g}} &= \frac{T}{2\pi} \\ \Rightarrow \frac{l}{g} &= \left(\frac{T}{2\pi}\right)^{2} \\ \Rightarrow l &= \frac{gT^{2}}{4\pi^{2}}. \end{aligned} \]
Substituting \(T = 2\ \text{s}\) and \(g = 9.8\ \text{m s}^{-2}\),
\[ \begin{aligned} l &= \frac{9.8 \times (2)^{2}}{4\pi^{2}} \\ &= \frac{9.8 \times 4}{4\pi^{2}} \\ &= \frac{39.2}{4\pi^{2}} \\ &\approx 0.994\ \text{m}. \end{aligned} \]
Hence, the length of a simple pendulum that ticks seconds is approximately \[ l \approx 1.0\ \text{m}. \]
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