2023 A/L Physics Past Paper · Question 35 (English Medium)

Answer: (2) $392\,\text{Hz}$

What the question asks

What we know:

The pipe is open at both ends. At the temperature $T$ it resonates at $f_1 = 400\,\text{Hz}$.

On another day the speed of sound is $2\%$ lower. The pipe is the same pipe.

We want: the frequency $f_2$ at which it resonates on that day.

What each letter means:

$f$ ($\text{Hz}$): the frequency.

$v$ ($\text{m s}^{-1}$): the speed of sound.

$\lambda$ ($\text{m}$): the wavelength.

The numbers $1$ and $2$ tell the two days apart. Day $1$ is at $T$.

Relevant theory

Even in a tube open at both ends, a sound wave admitted from one end when reaching the other open end gets reflected by a certain amount. Hence here too if the length of the tube could be adjusted to suit the wave length of the wave,

The relationship among the wave velocity v, frequency f and the wavelength λ, can be written in the form of v = f λ. This equation can be applied to all types of waves.

The first quote says that the length of the tube has to suit the wavelength of the wave for resonance. The second quote says that speed, frequency and wavelength go together as $v = f\lambda$.

For example, think of a flute. The flute is the same flute every day. On a warm afternoon the notes come out a little higher, and on a cold morning a little lower.

The sound travels faster in warm air. The flute still makes the same wavelength, so the faster sound gives a higher note.

We use the same idea here, with a slower sound.

Required equations

$v = f\lambda$

$f = \dfrac{v}{\lambda}$

$f \propto v$

Reasoning and calculation

What stays the same: the pipe. Its length does not change, so the wavelength that fits it for resonance is the same on both days.

What changes: the speed of sound. So the frequency changes with it.

Write it for each day, then divide.

$$f_1 = \dfrac{v_1}{\lambda} \qquad f_2 = \dfrac{v_2}{\lambda}$$

$$\dfrac{f_2}{f_1} = \dfrac{v_2}{v_1}$$

The $\lambda$ cancels, because it is the same on both days.

The speed is $2\%$ lower, so it is $98\%$ of the first speed.

$$\dfrac{v_2}{v_1} = 0.98$$

So the frequency is $98\%$ of the first one.

$$f_2 = 0.98 \times 400 = 392\,\text{Hz}$$

$f * r$

Reading the options

Now match it with the five options.

(1) $384\,\text{Hz}$ is $4\%$ lower. It does not match.

(2) $392\,\text{Hz}$ is $2\%$ lower. This matches.

(3) $396\,\text{Hz}$ is $1\%$ lower. It does not match.

(4) $408\,\text{Hz}$ is higher. It does not match.

(5) $416\,\text{Hz}$ is higher. It does not match.

Checks

Dimension check:

$\dfrac{f_2}{f_1}$ is a frequency divided by a frequency, and $\dfrac{v_2}{v_1}$ is a speed divided by a speed. Both are plain numbers, so they can be put equal.

Does it make sense?

A slower sound gives a lower frequency in the same pipe, so the answer must be less than $400\,\text{Hz}$.

A change of $2\%$ of $400\,\text{Hz}$ is $8\,\text{Hz}$, and $400 - 8 = 392\,\text{Hz}$. The same answer.

Conclusion

$f_2 = 392\,\text{Hz}$

The pipe resonates at $392\,\text{Hz}$.

The answer is (2).

The quotes give the length condition for an open pipe and $v = f\lambda$. That the wavelength that fits the pipe stays the same when only the speed of sound changes, so that $f$ goes with $v$, is our reasoning from them. We ignore any change in the length of the pipe with temperature.

Teacher's note

For the same pipe the wavelength stays the same, so the frequency goes with the speed of sound: $\dfrac{f_2}{f_1} = \dfrac{v_2}{v_1}$.

A speed that is $2\%$ lower gives a frequency that is $2\%$ lower.

Exam tip