Question:

For the system given below, find time period of oscillation?

Updated On: Jul 28, 2022
  • $\frac{10}{\sqrt{3}}$
  • $10\sqrt{3}$
  • $\frac{20}{\sqrt{3}}$
  • $20\sqrt{3}$
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The Correct Option is C

Solution and Explanation

By energy method $\frac{1}{2}Kx^{2}+\frac{1}{2}mv^{2}+\frac{1}{2}l\omega^{2}=C$ $\frac{1}{2}.K.2x.\frac{dx}{dt}+\frac{1}{2}.m.2v \frac{dv}{dt}+\frac{1}{2}I.+\frac{2v}{r^{2}} \frac{dv}{dt}=0$ $Kxv +\frac{M}{4}va+\frac{M}{2}a.v=0$ $-Kx=\frac{3Ma}{4}$ $a=-\frac{4K}{3M}.x=-\omega^{2}x$ $\omega^{2}=\frac{4K}{3M}$ $\omega=\sqrt{\frac{4K}{3M}}=\sqrt{\frac{4\times100}{3\times1}}=\frac{20}{\sqrt{3}}$ sec
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Concepts Used:

Energy In Simple Harmonic Motion

We can note there involves a continuous interchange of potential and kinetic energy in a simple harmonic motion. The system that performs simple harmonic motion is called the harmonic oscillator.

Case 1: When the potential energy is zero, and the kinetic energy is a maximum at the equilibrium point where maximum displacement takes place.

Case 2: When the potential energy is maximum, and the kinetic energy is zero, at a maximum displacement point from the equilibrium point.

Case 3: The motion of the oscillating body has different values of potential and kinetic energy at other points.