A simple harmonic oscillator has a time period of 10 seconds. Which equation rotates its acceleration 'a' and displacement 'x'?
Correct answer: C. a = -(2π/10)2 x
- A. a = -2 x
- B. a = -(20π)x
- C. a = -(2π/10)2 x
- D. a = -(20π)2 x
Explanation
The motion of a simple harmonic oscillator can be described by the following equation:a = -ω2 * xwhere 'a' is the acceleration, 'x' is the displacement, and ω is the angular frequency of the oscillator.The time period of the oscillator is given by:T = 2π/ωwhere 'T' is the time period.In this case, the time period is given as 10 seconds. Therefore, we can find the angular frequency as:ω = 2π/ T = 2π /10 =π/5Substituting this value of ω in the equation of motion, we get:a = - (π /5)2 * xSimplifying, we get:a = - (2π /10)2 * xTherefore, the correct option is C) a = - (2π / 10)2 * x
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About Simple Harmonic Motion
Simple harmonic motion is oscillation in which acceleration is directly proportional to displacement and directed toward the equilibrium position. Work includes displacement, velocity, acceleration, phase, period, frequency, amplitude and energy, with applications to springs and simple pendulums. The restoring force and the conditions for SHM distinguish it from general periodic motion.
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