A simple harmonic oscillator a time period of 10s, write an equation relates it's acceleration and displacement x

Physics

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Answer

a=−π225xa = -\frac{\pi^2}{25} x

Step-by-step solution

3 steps

  1. The general equation for acceleration aa in simple harmonic motion (SHM) is given by

    a=−ω2xa = -\omega^2 x

    where ω\omega is the angular frequency and xx is the displacement.

  2. Calculate the angular frequency ω\omega using the given time period T=10T = 10 s.

    ω=2πT\omega = \frac{2\pi}{T} ω=2π10 s\omega = \frac{2\pi}{10 \text{ s}} ω=π5 rad/s\omega = \frac{\pi}{5} \text{ rad/s}

  3. Substitute the value of ω\omega into the acceleration equation.

    a=−(π5)2xa = -\left(\frac{\pi}{5}\right)^2 x a=−π225xa = -\frac{\pi^2}{25} x
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