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Oscillations - Velocity and Acceleration in SHM

Grade 11CBSEPhysics

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

🔑Concepts

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Simple Harmonic Motion (SHM) is defined as a periodic motion where the acceleration of the particle is directly proportional to its displacement from the mean position and is always directed towards that mean position.

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The velocity v(t)v(t) of a particle in SHM is the first derivative of displacement x(t)x(t) with respect to time. It is represented as v(t)=dxdtv(t) = \frac{dx}{dt}.

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The acceleration a(t)a(t) of a particle in SHM is the first derivative of velocity v(t)v(t) or the second derivative of displacement x(t)x(t) with respect to time: a(t)=dvdt=d2xdt2a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2}.

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At the mean position (x=0x = 0), the velocity is at its maximum magnitude (v=ωAv = \omega A) and the acceleration is zero.

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At the extreme positions (x=±Ax = \pm A), the velocity is zero and the acceleration is at its maximum magnitude (a=ω2Aa = \omega^2 A).

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There is a phase difference of π2\frac{\pi}{2} between displacement and velocity, and a phase difference of π\pi between displacement and acceleration.

📐Formulae

x(t)=Acos⁡(ωt+ϕ)x(t) = A \cos(\omega t + \phi) or x(t)=Asin⁡(ωt+ϕ)x(t) = A \sin(\omega t + \phi) displacements

v(t)=−ωAsin⁡(ωt+ϕ)v(t) = -\omega A \sin(\omega t + \phi) (velocity as a function of time)

v=±ωA2−x2v = \pm \omega \sqrt{A^2 - x^2} (velocity as a function of position)

a(t)=−ω2Acos⁡(ωt+ϕ)a(t) = -\omega^2 A \cos(\omega t + \phi) (acceleration as a function of time)

a=−ω2xa = -\omega^2 x (acceleration as a function of position)

vmax=ωAv_{max} = \omega A

amax=ω2Aa_{max} = \omega^2 A

ω=2πT=2πf\omega = \frac{2\pi}{T} = 2\pi f

💡Examples

Problem 1:

A body oscillates with SHM according to the equation x=5cos⁡(2πt+π4)x = 5 \cos(2\pi t + \frac{\pi}{4}), where xx is in meters and tt is in seconds. Calculate the velocity and acceleration of the body at t=1.5 st = 1.5 \text{ s}.

Solution:

  1. Identify parameters: Amplitude A=5 mA = 5 \text{ m}, angular frequency ω=2π rad/s\omega = 2\pi \text{ rad/s}, phase constant ϕ=π4\phi = \frac{\pi}{4}.
  2. Velocity formula: v(t)=−ωAsin⁡(ωt+ϕ)v(t) = -\omega A \sin(\omega t + \phi). At t=1.5 st = 1.5 \text{ s}: v=−(2π)(5)sin⁡(2π×1.5+π4)=−10πsin⁡(3π+π4)=−10π(−sin⁡π4)=10π2≈22.21 m/sv = -(2\pi)(5) \sin(2\pi \times 1.5 + \frac{\pi}{4}) = -10\pi \sin(3\pi + \frac{\pi}{4}) = -10\pi (-\sin \frac{\pi}{4}) = \frac{10\pi}{\sqrt{2}} \approx 22.21 \text{ m/s}.
  3. Acceleration formula: a=−ω2xa = -\omega^2 x. First, find xx at t=1.5t = 1.5: x=5cos⁡(3π+π4)=5(−cos⁡π4)=−52 mx = 5 \cos(3\pi + \frac{\pi}{4}) = 5 (-\cos \frac{\pi}{4}) = -\frac{5}{\sqrt{2}} \text{ m}. a=−(2π)2(−52)=20π22≈139.57 m/s2a = -(2\pi)^2 (-\frac{5}{\sqrt{2}}) = \frac{20\pi^2}{\sqrt{2}} \approx 139.57 \text{ m/s}^2.

Explanation:

We use the time-dependent equations for velocity and acceleration derived from the displacement equation. The phase (3π+π4)(3\pi + \frac{\pi}{4}) falls in the third quadrant where both sine and cosine are negative.

Problem 2:

A particle executing SHM has a maximum velocity of 1 m/s1 \text{ m/s} and a maximum acceleration of 1.57 m/s21.57 \text{ m/s}^2. Determine the time period TT and the amplitude AA.

Solution:

  1. Given: vmax=ωA=1v_{max} = \omega A = 1 and amax=ω2A=1.57a_{max} = \omega^2 A = 1.57.
  2. Divide amaxa_{max} by vmaxv_{max}: ω2AωA=1.571⇒ω=1.57 rad/s\frac{\omega^2 A}{\omega A} = \frac{1.57}{1} \Rightarrow \omega = 1.57 \text{ rad/s}.
  3. Since ω=π2\omega = \frac{\pi}{2} (approx 1.57≈3.1421.57 \approx \frac{3.14}{2}), we find T=2πω=2π1.57≈4 sT = \frac{2\pi}{\omega} = \frac{2\pi}{1.57} \approx 4 \text{ s}.
  4. Find AA from vmax=ωAv_{max} = \omega A: 1=(1.57)A⇒A=11.57≈0.637 m1 = (1.57) A \Rightarrow A = \frac{1}{1.57} \approx 0.637 \text{ m}.

Explanation:

By taking the ratio of maximum acceleration to maximum velocity, we isolate the angular frequency ω\omega. Once ω\omega is known, the time period and amplitude are easily calculated using standard SHM relations.

Velocity and Acceleration in SHM - Revision Notes & Key Formulas | CBSE Class 11 Physics