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I am studying SHM in my physics class right now and I often get confused with the formula for displacement.

Sometimes I see the formula written as $x=A\sin(\omega t)$ and sometimes I see it written as $x = A\cos(\omega t)$. So my question what exactly is the formula for displacement in Simple Harmonic Motion?

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Simple harmonic motion response is sinusoidal with a frequency of $\omega$, so the general case is expressed as $$x=A\sin(\omega t+\phi)=A\cos(\omega t-\pi/2+\phi)$$ or $$x=B\sin(\omega t)+C\cos(\omega t),$$

where $A$ is the amplitude, $\phi$ is the so-called phase angle (whose negative, $-\phi$, is called the phase delay), and $B$ and $C$ are constants related to the amplitude and phase angle through

$$A^2=B^2+C^2$$ and $$B=\cos\phi\qquad C=\sin\phi\qquad\phi=\tan^{-1}(C/B)$$ (where the arctangent is limited between $-\pi/2$ and $\pi/2$). You can find these constants from the boundary conditions. If $x=0$ at $t=0$ (i.e., the displacement is zero at time zero), for example, then we have simply $x=A\sin(\omega t)$. If instead $\dot x=dx/dt=0$ at $t=0$ (i.e., the speed is zero at time zero), then $x=A\cos(\omega t)$. Does this make sense?

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  • $\begingroup$ You're saying "If ๐‘ฅ=0 at ๐‘ก=0, for example, then we have simply ๐‘ฅ=๐ดsin(๐œ”๐‘ก). If instead ๐‘ฅห™=0 at ๐‘ก=0, then ๐‘ฅ=๐ดcos(๐œ”๐‘ก)." I don't quite understand. You gave 2 values for x in the same condition? $\endgroup$ Jan 20, 2021 at 19:25
  • $\begingroup$ I've edited the answer to provide more detail. $\endgroup$ Jan 20, 2021 at 19:36

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