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Oscillations are repetitive back-and-forth motions around a stable central point called the equilibrium position. Imagine pushing a child on a swing: after you give a push, the swing moves forward and backward repeatedly, passing through the lowest point each time. This is a classic example of oscillatory motion.
Oscillations are fundamental in physics and engineering because many natural and man-made systems exhibit such behavior-ranging from vibrating guitar strings and pendulum clocks to electrical circuits and even the motion of atoms in solids. Understanding oscillations is crucial for solving problems in mechanics and for competitive exams where questions often test conceptual clarity and problem-solving skills.
Among oscillations, Simple Harmonic Motion (SHM) is a special and important type. It occurs when the restoring force acting on the oscillating object is directly proportional to its displacement from equilibrium and is always directed towards that equilibrium.
In simpler terms, the farther you pull or push the object from its resting position, the stronger the force pulling it back - like a spring that pulls harder the more it is stretched or compressed.
Consider a mass attached to a spring on a frictionless surface. When displaced by a distance \( x \) from equilibrium, the spring exerts a restoring force \( F \) given by Hooke's Law:
\[ F = -kx \]
Here, \( k \) is the spring constant, a measure of the spring's stiffness, and the negative sign indicates the force is opposite to the displacement direction.
According to Newton's second law, force equals mass times acceleration:
\[ F = m a = m \frac{d^2 x}{dt^2} \]
Equating the two expressions for force:
\[ m \frac{d^2 x}{dt^2} = -kx \]
Rearranging:
\[ \frac{d^2 x}{dt^2} + \frac{k}{m} x = 0 \]
This is a second-order differential equation describing SHM. Its general solution is:
\[ x(t) = A \cos(\omega t + \phi) \]
where:
In this diagram, the mass is displaced to the right by \( x \). The restoring force \( F \) points left, towards equilibrium. The velocity vector \( v \) points downward here, indicating direction of motion at that instant (depending on phase).
Energy in SHM continuously transforms between two forms:
The potential energy at displacement \( x \) is:
\[ PE = \frac{1}{2} k x^2 \]
The kinetic energy at velocity \( v \) is:
\[ KE = \frac{1}{2} m v^2 \]
Since velocity varies with time as the mass oscillates, energy shifts back and forth between PE and KE, but the total mechanical energy remains constant (assuming no friction or damping):
\[ E = PE + KE = \frac{1}{2} k A^2 = \text{constant} \]
In the graph above, the blue curve shows kinetic energy (KE) and the red curve shows potential energy (PE) as functions of displacement. At maximum displacement (\( x = \pm A \)), KE is zero and PE is maximum. At equilibrium (\( x=0 \)), PE is zero and KE is maximum.
In real systems, oscillations are rarely ideal. Two important phenomena modify oscillations:
graph TD A[Oscillations] --> B[Free Oscillations] B --> C[Undamped] B --> D[Damped] A --> E[Forced Oscillations] E --> F[Resonance]
Damped Oscillations: The displacement equation becomes:
\[ x(t) = A e^{-\gamma t} \cos(\omega' t + \phi) \]
where \( \gamma \) is the damping coefficient and \( \omega' \) is the damped angular frequency, slightly less than the natural frequency \( \omega \).
Forced Oscillations and Resonance: When an external force oscillates at frequency close to the system's natural frequency, the amplitude can increase dramatically, a phenomenon called resonance. This is important in engineering to avoid structural failures.
Step 1: Recall the formula for the period of a simple pendulum:
\[ T = 2\pi \sqrt{\frac{l}{g}} \]
Step 2: Substitute the values:
\[ T = 2\pi \sqrt{\frac{1.5}{9.8}} \]
\[ \sqrt{\frac{1.5}{9.8}} = \sqrt{0.153} \approx 0.391 \]
\[ T = 2 \times 3.1416 \times 0.391 \approx 2.46 \, s \]
Answer: The time period is approximately 2.46 seconds.
Step 1: Calculate angular frequency \( \omega \):
\[ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.5}} = \sqrt{400} = 20 \, rad/s \]
Step 2: Use the maximum speed formula:
\[ v_{max} = \omega A = 20 \times 0.1 = 2 \, m/s \]
Answer: The maximum speed is 2 m/s.
Step 1: Calculate total mechanical energy \( E \):
\[ E = \frac{1}{2} k A^2 = \frac{1}{2} \times 100 \times (0.05)^2 = 0.125 \, J \]
Step 2: At maximum displacement \( x = A \), velocity \( v = 0 \), so:
\[ PE = \frac{1}{2} k A^2 = 0.125 \, J \]
\[ KE = 0 \]
Total energy: \( PE + KE = 0.125 + 0 = 0.125 \, J \)
Step 3: At equilibrium \( x = 0 \), potential energy is zero and velocity is maximum:
\[ PE = 0 \]
Calculate \( \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{100}{0.3}} \approx 18.26 \, rad/s \)
\[ v_{max} = \omega A = 18.26 \times 0.05 = 0.913 \, m/s \]
\[ KE = \frac{1}{2} m v_{max}^2 = \frac{1}{2} \times 0.3 \times (0.913)^2 \approx 0.125 \, J \]
Total energy: \( PE + KE = 0 + 0.125 = 0.125 \, J \)
Answer: Total mechanical energy remains constant at 0.125 J.
Step 1: Use the damped amplitude relation:
\[ A(t) = A_0 e^{-\gamma t} \]
Step 2: Substitute known values:
\[ 0.1 = 0.2 \times e^{-\gamma \times 5} \]
\[ \frac{0.1}{0.2} = e^{-5\gamma} \Rightarrow 0.5 = e^{-5\gamma} \]
Step 3: Take natural logarithm on both sides:
\[ \ln(0.5) = -5\gamma \Rightarrow -0.693 = -5\gamma \]
\[ \gamma = \frac{0.693}{5} = 0.1386 \, s^{-1} \]
Answer: The damping coefficient is approximately 0.139 s\(^{-1}\).
Step 1: Calculate natural angular frequency:
\[ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{50}{2}} = \sqrt{25} = 5 \, rad/s \]
Step 2: Convert angular frequency to frequency:
\[ f = \frac{\omega}{2\pi} = \frac{5}{6.283} \approx 0.796 \, Hz \]
Step 3: Significance: At this frequency, if the external driving force oscillates, the system's amplitude will increase significantly due to resonance, which can cause large oscillations and possible damage.
Answer: Resonance frequency is approximately 0.796 Hz.
When to use: When identifying or verifying SHM problems.
When to use: In competitive exams where time is limited.
When to use: When calculating pendulum periods.
When to use: When analyzing damped oscillation problems.
When to use: During formula recall in exams.
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