Problem Solving Simple Harmonic Motion
Concepts to remember on Problem-Solving Simple Harmonic Motion for JEE and CBSE board exams:
Simple Harmonic Motion (SHM):
- Periodic motion where the restoring force is directly proportional to the negative displacement.
- Examples: Pendulum, spring-mass system, and oscillating bodies.
- Motion takes place along a straight line.
Characteristics of SHM:
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Displacement (x): A sinusoidal function of time (t) with amplitude (A) and angular frequency (ω).
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x(t) = A cos(ωt + Φ) or x(t) = A sin(ωt + Φ)
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Velocity (v): Also sinusoidal, 90° out of phase with displacement.
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v(t) = -ωA sin(ωt + Φ) or v(t) = ωA cos(ωt + Φ)
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Acceleration (a): Sinusoidal, 180° out of phase with displacement and proportional to (-Aω²).
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a(t) = -ω²A cos(ωt + Φ) or a(t) = -ω²A sin(ωt + Φ)
Equations of SHM:
- Displacement: x = A cos(ωt + ∅), where ∅ is the phase constant.
- Velocity: v = -Aω sin(ωt + ∅).
- Acceleration: a = -Aω² cos(ωt + ∅).
Time Period (T):
- Time taken for one complete oscillation.
- T = 2π√(m/k), where m is the mass of the oscillating object, and k is the spring constant.
Frequency (f):
- Number of oscillations per second.
- f = 1/T = ω/2π.
Angular Frequency (ω):
- Measures the rate of change of phase angle.
- ω = 2πf = √(k/m).
Phase:
- Describes the position of an oscillating particle within its cycle.
- Phase difference: Describes the difference in phase between two oscillations.
Energy in SHM:
- Total energy (E) is constant and the sum of kinetic energy (K) and potential energy (U).
- K = ½kA², and U = ½kA² cos²(ωt + ∅).
- E = ½kA².
Resonance:
- Occurs when the frequency of an applied periodic force matches the natural frequency of the system.
- Causes maximum amplitude and energy transfer.