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Describe simple examples of free oscillations such as the simple pendulum, spring-mass system and torsional pendulum.

Physics

ScanSolve AI Answer

Answer

a=−ω2xa = -\omega^2 x

Step-by-step solution

4 steps
  1. Start with the given displacement equation for simple harmonic motion.

    The displacement xx of a simple oscillator as a function of time tt is given by:

    x=x0sin⁡(ωt)x = x_0 \sin(\omega t)

    where x0x_0 is the amplitude and ω\omega is the angular frequency.

  2. Differentiate the displacement equation with respect to time to find the velocity.

    Velocity vv is the first derivative of displacement with respect to time:

    v=dxdtv = \frac{dx}{dt} v=ddt(x0sin⁡(ωt))v = \frac{d}{dt} (x_0 \sin(\omega t))

    Using the chain rule, ddt(sin⁡(at))=acos⁡(at)\frac{d}{dt} (\sin(at)) = a \cos(at):

    v=x0ωcos⁡(ωt)v = x_0 \omega \cos(\omega t)
  3. Differentiate the velocity equation with respect to time to find the acceleration.

    Acceleration aa is the first derivative of velocity with respect to time (or the second derivative of displacement):

    a=dvdta = \frac{dv}{dt} a=ddt(x0ωcos⁡(ωt))a = \frac{d}{dt} (x_0 \omega \cos(\omega t))

    Using the chain rule, ddt(cos⁡(at))=−asin⁡(at)\frac{d}{dt} (\cos(at)) = -a \sin(at):

    a=x0ω(−ωsin⁡(ωt))a = x_0 \omega (-\omega \sin(\omega t)) a=−x0ω2sin⁡(ωt)a = -x_0 \omega^2 \sin(\omega t)
  4. Substitute the expression for xx back into the acceleration equation.

    From Step 1, we know that x=x0sin⁡(ωt)x = x_0 \sin(\omega t). We can substitute this into the expression for aa from Step 3:

    a=−ω2(x0sin⁡(ωt))a = -\omega^2 (x_0 \sin(\omega t)) a=−ω2xa = -\omega^2 x

    This proves that for simple oscillations, the acceleration is directly proportional to the negative of the displacement, which is the defining characteristic of simple harmonic motion.

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