2018 A/L Physics Past Paper · Question 17 (English Medium)

Answer: (2) 1, 1, 2π, 4π, 8π²

What the question asks

We know: the x – t graph of an object in simple harmonic motion.

We want: $T$, $f$, $\omega$, $v_{max}$ and $a_{max}$.

Then we find the row of the table that has all five.

Reading the options

The table has five columns.

$T$ is in s. $f$ is in Hz. $\omega$ is in $s^{-1}$.

$v_{max}$ and $a_{max}$ are given in units of $10^{-2}$.

Relevant theory

If the acceleration of a moving body is always directed towards an equilibrium point, its magnitude being directly proportional to the distance from the same point, then that motion of the body is called simple harmonic motion.

Required equations

$f = \frac{1}{T},\quad \omega = 2\pi f$

$v_{max} = A\omega$

$a_{max} = A\omega^2$

Reasoning and calculation

Read the axes first.

$t$ is in seconds.

$x$ is in units of $10^{-2}$ m.

The curve starts at 0. It crosses zero at 0.5 s, 1 s, 1.5 s, 2 s and 2.5 s.

The highest value is $+2 \times 10^{-2}$ m.

The lowest value is $-2 \times 10^{-2}$ m.

So the amplitude is $A = 2 \times 10^{-2}$ m.

The curve goes up, comes down, and is back at the start, moving up again, at $t = 1$ s.

So one full oscillation takes 1 s.

$1$

$T$

$2\times10^{-2}$

$A$

$T$ is the period. $A$ is the amplitude.

From these two we get all the rest.

$f = 1/T$

$2\pi$

$\omega = 2\pi f$

$4\pi\times10^{-2}$

$v_{max} = 2\times10^{-2}\times2\pi$

$8\pi^2\times10^{-2}$

$a_{max} = 2\times10^{-2}\times(2\pi)^2$

So the five values are $1$, $1$, $2\pi$, $4\pi$ and $8\pi^2$.

They match row (2) of the table.

Checks

$A * (2 * pi / T)^2$

Check the units.

$$A\omega = (m)(s^{-1}) = m\,s^{-1}$$

That is the unit of a speed.

$$A\omega^{2} = (m)(s^{-2}) = m\,s^{-2}$$

That is the unit of an acceleration. Both match the table.

Conclusion

$T = 1$ s. $f = 1$ Hz. $\omega = 2\pi\ s^{-1}$.

$$v_{max} = A\omega = 4\pi \times 10^{-2}\ \text{m s}^{-1}$$

$$a_{max} = A\omega^2 = 8\pi^2 \times 10^{-2}\ \text{m s}^{-2}$$

The answer is option (2).

Taking the time between two zero crossings as T.

That is only half a swing, $T/2$.

A full oscillation ends when the object is back at the same place, moving the same way.

Explore

The mass on the spring moves up and down. Its pen draws the x – t graph.

From $x = 0$ back to $x = 0$ takes 0.5 s. That is only half a swing.

Moving the same way and back to $x = 0$ again takes 1 s. So $T = 1$ s.

In the middle the mass moves fastest. At the ends it stops and turns back. There the acceleration is largest.