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Oscillations - Simple Harmonic Motion and Uniform Circular Motion

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Simple Harmonic Motion (SHM) is a periodic motion where the restoring force is directly proportional to the displacement from the equilibrium position and acts in the opposite direction: F=−kxF = -kx.

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SHM can be understood as the projection of Uniform Circular Motion (UCM) on any diameter of the circle of reference. The radius of the circle is equal to the amplitude AA of the oscillation.

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The angular speed ω\omega of the particle in UCM becomes the angular frequency of the SHM.

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The phase (ωt+ϕ)(\omega t + \phi) of a vibrating particle at any instant is the state of its motion (both position and direction) at that instant. ϕ\phi is the initial phase or epoch.

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Total energy in SHM is conserved (in the absence of friction). It is the sum of Potential Energy U=12kx2U = \frac{1}{2}kx^2 and Kinetic Energy K=12mv2K = \frac{1}{2}mv^2.

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Velocity is maximum at the mean position (x=0x = 0) and zero at the extreme positions (x=±Ax = \pm A). Acceleration is maximum at the extreme positions and zero at the mean position.

📐Formulae

x(t)=Acos⁡(ωt+ϕ)x(t) = A \cos(\omega t + \phi)

ω=2πT=2πf=km\omega = \frac{2\pi}{T} = 2\pi f = \sqrt{\frac{k}{m}}

v(t)=−ωAsin⁡(ωt+ϕ)v(t) = -\omega A \sin(\omega t + \phi)

v=±ωA2−x2v = \pm \omega \sqrt{A^2 - x^2}

a(t)=−ω2Acos⁡(ωt+ϕ)=−ω2xa(t) = -\omega^2 A \cos(\omega t + \phi) = -\omega^2 x

U=12mω2x2U = \frac{1}{2} m \omega^2 x^2

K=12mω2(A2−x2)K = \frac{1}{2} m \omega^2 (A^2 - x^2)

Etotal=12mω2A2=12kA2E_{total} = \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} k A^2

💡Examples

Problem 1:

A particle executes SHM with an amplitude of 0.1 m0.1 \text{ m} and a time period of 2 s2 \text{ s}. Find the maximum velocity and maximum acceleration of the particle.

Solution:

Given amplitude A=0.1 mA = 0.1 \text{ m} and T=2 sT = 2 \text{ s}. First, calculate angular frequency: ω=2πT=2π2=π rad/s\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \text{ rad/s} Now, maximum velocity is: vmax=Aω=0.1×π=0.1π m/s≈0.314 m/sv_{max} = A\omega = 0.1 \times \pi = 0.1\pi \text{ m/s} \approx 0.314 \text{ m/s} Maximum acceleration is: amax=Aω2=0.1×π2=0.1π2 m/s2≈0.987 m/s2a_{max} = A\omega^2 = 0.1 \times \pi^2 = 0.1\pi^2 \text{ m/s}^2 \approx 0.987 \text{ m/s}^2

Explanation:

In SHM, maximum velocity occurs at the equilibrium position where v=Aωv = A\omega, and maximum acceleration occurs at the extreme points where a=ω2Aa = \omega^2 A.

Problem 2:

A 2 kg2 \text{ kg} mass is attached to a spring with k=200 N/mk = 200 \text{ N/m}. If the mass is displaced by 5 cm5 \text{ cm}, find the total energy of the system.

Solution:

Given m=2 kgm = 2 \text{ kg}, k=200 N/mk = 200 \text{ N/m}, and A=5 cm=0.05 mA = 5 \text{ cm} = 0.05 \text{ m}. The total energy EE is given by: E=12kA2E = \frac{1}{2} k A^2 E=12×200×(0.05)2E = \frac{1}{2} \times 200 \times (0.05)^2 E=100×0.0025E = 100 \times 0.0025 E=0.25 JE = 0.25 \text{ J}

Explanation:

The total energy of a spring-mass system in SHM is independent of time and is equal to the maximum potential energy at the extreme displacement.