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1. Which of the following condition is sufficient for the simple harmonic motion?

  • a = ky
  • a = ky2
  • a = -ky
  • a = -ky2

Where ‘a’ – acceleration, y – displacement

2. Prove that simple harmonic motion is the projection of uniform circular motion on any diameter of the circle.

3. Represent graphically the variations of potential energy, kinetic energy and total energy as a function of position ‘x’ for a linear harmonic oscillator. Explain the graph.

1 Answer

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1. a = -ky

2.

Consider a particle moving along the circumference of a circle of radius ‘a’ and center O, with uniform angular velocity w. AB and CD are two mutually perpendicular diameters along X and Y axis. At time t = 0.

let the particle be at P0 so that ∠P0OB = Φ. After time ‘t’ second, let the particle reach P so that ∠POP0 = ω t. N is the foot of the perpendicular drawn from P on the diameter CD.

Similarly M is the foot of the perpendicular drawn from P to the diameter AB. When the particle moves along the circumference of the circle, the foot of the perpendicular executes to and for motion along the diameter CD or AB with O as the mean position.

From the right angle triangle O MP, we get

Cos (ωt + Φ) = OMOP

∴ OM = OPcos(ωt + Φ)

X= a cos (ωt + Φ) _______(1)

Similarly, we get

Sin (ωt + Φ) = ya (or)

Y = a sin (ωt + Φ) _______(2)

Equation (1) and (2) are similar to equations of S.H.M. The equation(1) and (2) shows that the projection of uniform circular motion on any diameter is S.H.M.

3. KE = PE

\(\frac{1}{2}\)2(a2 – x2) = \(\frac{1}{2}\) = mω2x2

Solving we get, x = \(\frac{a^2}{√2}\)

where a is the amplitude of oscillation.

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