2021 A/L Physics Past Paper · Question 13 (English Medium)

Answer: (1) Graph pair (1)

What the question asks

We want: the graphs of the displacement $x$ and the velocity $v$ of $P$ against the time $t$.

$P$ is the foot of the perpendicular from $A$ to the $X$ axis.

Relevant theory

The foot of the perpendicular projection (M) from the moving body on to this diameter moves from Z to C, and passing through Z again and moving to F returns to Z.

Required equations

$x = R\cos(\omega t + \varphi)$

$v = \dfrac{dx}{dt} = -R\omega\sin(\omega t + \varphi)$

Reasoning and calculation

Where does $P$ start?

In the figure $A$ is above the $X$ axis and to the right of the centre.

So $P$ starts on the positive side. $x$ starts positive, not at zero.

Which way does $P$ go?

$A$ turns anticlockwise, towards the $Y$ axis. So $P$ moves to the left, towards the centre.

So $x$ falls from its starting value. It reaches zero, goes negative to the other end, and then comes back.

Now the velocity. It is the slope of the $x$-$t$ graph.

At first $x$ is falling, so $v$ is negative.

Where $x$ crosses zero going down, the slope is steepest. $v$ is at its most negative.

At the lowest point of $x$ the slope is zero. $v = 0$.

Where $x$ crosses zero going up, the slope is steepest the other way. $v$ is at its largest positive value.

Line up the dashed lines.

Dashed line 1: $x = 0$ going down, and $v$ is at its lowest.

Dashed line 2: $x$ is at its lowest, and $v = 0$.

Dashed line 3: $x = 0$ going up, and $v$ is at its highest.

Only the pair (1) has all of these.

Pair (2) starts at the centre. Pair (3) has $v$ copying $x$. Pairs (4) and (5) have no changing $v$.

The answer is pair (1).

Checks

$x * x + u * u$

Does it make sense?

When $P$ is at an end of its path it stops for a moment. At those points $v = 0$ and $x$ is at its largest or smallest.

When $P$ is at the centre it moves fastest. At those points $x = 0$ and $|v|$ is at its largest.

Conclusion

$\text{pair }(1)$

$x$ starts positive and falls. $v$ is the slope of $x$, so it starts negative and is zero when $x$ is at its lowest.

The answer is (1).

The velocity graph is the slope of the displacement graph.

Where $x$ is at a peak or a trough, $v = 0$. Where $x$ crosses zero, $|v|$ is largest.

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