In a simple pendulum of length 10 m, the string is initially kept horizontal and the bob is released. 10% of energy is lost till the bob reaches the lowest position. Then find the speed of the bob at the lowest position.
- \(6 \;m/s\)
- \(6\sqrt5 \;m/s\)
- \(7\sqrt5 \;m/s\)
- \(4\sqrt2 \;m/s\)
The Correct Option is B
Solution and Explanation
The correct option is (B): \(6\sqrt5 \;m/s\)
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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.