2022 A/L Physics Past Paper · Question 17 (English Medium)

Answer: (4) $\dfrac{2T}{\sqrt{3}}$

Given information

What we know:

The pendulum of length $l$ has period $T$.

The new pendulum has length $l' = 2l$.

The lift accelerates upward at $a = \dfrac{g}{2}$.

What the question asks

We want: the period $T'$ of the new pendulum inside the lift.

What each letter means:

$T$ ($\text{s}$): the period of the first pendulum.

$l$ ($\text{m}$): the length of the first pendulum.

$g$ ($\text{m s}^{-2}$): acceleration due to gravity.

$a$ ($\text{m s}^{-2}$): the upward acceleration of the lift.

$g'$ ($\text{m s}^{-2}$): the effective gravity felt inside the lift.

Relevant theory

T depends only on the length of the pendulum. At the location of the pendulum, g is considered to be a constant.

T = 2π√(l/g)

A force of 1 newton (N), is the force which would give a 1 kg mass an acceleration of 1 m s^-2.

For example, stand on a bathroom scale inside a lift that starts to go up.

You feel heavier and the scale reads more. The floor has to push you up harder than your weight, because it must also speed you up.

A pendulum bob feels the same. The string has to pull the bob up harder than its weight. Inside the lift it is as if gravity had become stronger.

The bob stays at rest relative to the lift. So by Newton's second law, the pull of the string minus the weight gives the mass times the acceleration of the lift.

$$T_s - mg = ma$$

In words: the string pulls with $m(g + a)$, so the bob behaves as if $g$ were replaced by $g + a$.

$$g' = g + a$$

In words: when the lift accelerates upward, the effective gravity is the sum of $g$ and $a$.

Required equations

$T = 2\pi\sqrt{\dfrac{l}{g}}$

$g' = g + a$

$T' = 2\pi\sqrt{\dfrac{l'}{g'}}$

Applying it to this question

Everything here is a ratio, so we divide the two periods. The $2\pi$ cancels.

The units cancel too, so the ratio has no unit.

$1.5$

$g' = g + \dfrac{g}{2} = \dfrac{3g}{2}$

$2$

$\dfrac{l'}{l} = 2$

$1.3333333$

$\left(\dfrac{T'}{T}\right)^2 = \dfrac{l'}{l} \times \dfrac{g}{g'} = 2 \times \dfrac{2}{3}$

$1.1547005$

$\dfrac{T'}{T} = \sqrt{\dfrac{4}{3}} = \dfrac{2}{\sqrt{3}}$

Reasoning and calculation

The new period is $T' = \dfrac{2T}{\sqrt{3}}$.

That is option (4).

Checks

$(2 * l) / l * g / (1.5 * g)$

Does it make sense?

The doubled length makes the swing slower, by $\sqrt{2}$. The stronger effective gravity makes it faster, by $\sqrt{\dfrac{2}{3}}$.

The slow-down wins: $\sqrt{2} \times \sqrt{\dfrac{2}{3}} = \dfrac{2}{\sqrt{3}}$, about $1.15$. So $T'$ is a little more than $T$.

Conclusion

$T' = \dfrac{2T}{\sqrt{3}}$

The period in the lift is $T' = 2\pi\sqrt{\dfrac{2l}{3g/2}} = \dfrac{2T}{\sqrt{3}}$.

The answer is (4).

When a lift accelerates upward, use $g + a$ in place of $g$. When it accelerates downward, use $g - a$.

Divide the two periods, so the $2\pi$ and the units drop out.

Exam tip