Oscillations: Question 10

Worked solution

Part (a): Period and values at t = 0

The period is related to the angular frequency by: $T=\frac{2\pi}{\omega}=\frac{2\pi}{4.0}=1.5708\text{ s}$

Recompute as a check, using $\omega=2\pi/T$ rearranged: $T=2\pi/4.0=6.2832/4.0=1.5708\text{ s}$, the same value. So $T=1.57\text{ s}$ (3 s.f.).

At $t=0$, using $x_0=5.0\text{ cm}=0.050\text{ m}$: $x=x_0\cos(0)=x_0=0.050\text{ m}$

Differentiating $x=x_0\cos\omega t$ gives the velocity $v=-x_0\omega\sin\omega t$; at $t=0$, $\sin(0)=0$, so: $v=-x_0\omega\sin(0)=0$

The acceleration is $a=-\omega^2x$; at $t=0$, $x=x_0$, so: $a=-\omega^2x_0=-(4.0)^2\times0.050=-16\times0.050=-0.800\text{ m s}^{-2}$

So at $t=0$: $x=0.050\text{ m}$ (maximum positive displacement), $v=0$, $a=-0.800\text{ m s}^{-2}$ (maximum magnitude, directed back toward the centre).

Part (b): Values at t = T/4

$T/4=1.5708/4=0.3927\text{ s}$. The phase angle is $\omega t=4.0\times0.3927=1.5708\text{ rad}=\pi/2$ exactly (since $T/4$ was defined directly from $\omega$).

$x=x_0\cos(\pi/2)=0.050\times0=0$ $v=-x_0\omega\sin(\pi/2)=-0.050\times4.0\times1=-0.200\text{ m s}^{-1}$ $a=-\omega^2x=-16\times0=0$

Recompute the velocity as a check, using $v=\pm\omega\sqrt{x_0^2-x^2}$: at $x=0$, $\sqrt{x_0^2-0}=x_0=0.050\text{ m}$, so $|v|=\omega x_0=4.0\times0.050=0.200\text{ m s}^{-1}$. The same magnitude, and negative because P is moving in the negative direction at this point in the cycle.

So at $t=T/4$: $x=0$, $v=-0.200\text{ m s}^{-1}$ (maximum magnitude, since the particle passes through the centre here), $a=0$.

Part (c): Values at t = T/2

$T/2=1.5708/2=0.7854\text{ s}$; $\omega t=4.0\times0.7854=3.1416\text{ rad}=\pi$ exactly.

$x=x_0\cos(\pi)=0.050\times(-1)=-0.050\text{ m}$ $v=-x_0\omega\sin(\pi)=-0.050\times4.0\times0=0$ $a=-\omega^2x=-16\times(-0.050)=+0.800\text{ m s}^{-2}$

Recompute the acceleration as a check, using $a=-\omega^2x_0$ at the opposite extreme: since the particle is now at its negative extreme, $a$ should have the same magnitude as at $t=0$ but the opposite sign: $-(-0.800)=+0.800\text{ m s}^{-2}$, consistent.

So at $t=T/2$: $x=-0.050\text{ m}$ (maximum negative displacement), $v=0$, $a=+0.800\text{ m s}^{-2}$ (maximum magnitude, directed back toward the centre, i.e. now in the positive direction).

Part (d): Phase relationships

Velocity relative to displacement: Comparing the three time points, $v$ is zero exactly when $x$ is at an extreme ($t=0$ and $t=T/2$), and $v$ has its greatest magnitude exactly when $x=0$ ($t=T/4$). This means the v-t graph has the same repeating shape as the x-t graph, but shifted earlier along the time axis by a quarter of a period. The v-t graph therefore leads the x-t graph by $T/4$ (a quarter cycle, or $90°$/$\pi/2\text{ rad}$ of phase).

Acceleration relative to displacement: At every one of the three time points, $a$ has the opposite sign to $x$ (and is zero exactly when $x$ is zero): at $t=0$, $x>0$ and $a<0$; at $t=T/4$, both are zero; at $t=T/2$, $x<0$ and $a>0$. This is exactly what $a=-\omega^2x$ predicts. The a-t graph is a scaled, inverted (mirror-image) copy of the x-t graph at every instant, so the two graphs are always exactly out of phase (antiphase), a phase difference of half a period ($T/2$, or $180°$/$\pi\text{ rad}$).

Final answers

• (a) $T=\boxed{1.57}\text{ s}$; at $t=0$: $x=0.050\text{ m}$, $v=0$, $a=-0.800\text{ m s}^{-2}$
• (b) At $t=T/4$: $x=0$, $v=-0.200\text{ m s}^{-1}$, $a=0$
• (c) At $t=T/2$: $x=-0.050\text{ m}$, $v=0$, $a=+0.800\text{ m s}^{-2}$
• (d) v-t leads x-t by $T/4$ (quarter cycle); a-t is exactly antiphase with x-t (half-cycle, $T/2$, phase difference)