2022 A/L Physics Past Paper · Question 48 (English Medium)

Answer: (4) $76\ \text{Hz}$

Given information

What we know:

A tuning fork is sounded with a tube closed at one end.

With the air at $51\ ^\circ\text{C}$ the beat frequency is $4\ \text{Hz}$. With the air at $127\ ^\circ\text{C}$ the beat frequency is again $4\ \text{Hz}$.

The tube is the same, and it sounds the same overtone each time. The end correction is neglected.

The question gives $\sqrt{324} = 18$.

What the question asks

We want: the frequency of the tuning fork.

What each letter means:

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

$f_1$ ($\text{Hz}$): the note of the tube with air at $51\ ^\circ\text{C}$.

$f_2$ ($\text{Hz}$): the note of the tube with air at $127\ ^\circ\text{C}$.

$T_1$, $T_2$ ($\text{K}$): the two air temperatures on the kelvin scale.

$v_1$, $v_2$ ($\text{m s}^{-1}$): the speed of sound in the air at $T_1$ and $T_2$.

$\lambda$ ($\text{m}$): the wavelength of the sound in the tube.

Relevant theory

What is happening:

The air in the tube rings with a note of its own. The fork rings with its own note. When the two notes are close, we hear beats.

The tube and the overtone stay the same, so the wavelength of the sound in the tube stays the same.

Warm air carries sound faster. With the same wavelength, a faster sound means a higher note. So the hot tube has a higher note than the cooler tube.

The fork gives $4\ \text{Hz}$ beats with both. So the fork must sit exactly half way between the two tube notes, $4\ \text{Hz}$ above the low one and $4\ \text{Hz}$ below the high one.

λ₁/4 = l + e ⇒ λ₁ = 4(l + e)

3λ₂/4 = l + e ⇒ λ₂ = (4/3)(l + e)

The relationship among the wave velocity v, frequency f and the wavelength λ, can be written in the form of v = f λ.

dependent on temperature with the velocity increasing with increasing temperature.

v₁/v₂ = √(T₁/T₂)

v₀/v₃₀ = √(T₀/T₃₀) = √(273/303)

The number of times the sound reaches maximum in one second is called beat frequency,

The first two quotes give the wavelength in a closed tube. With the end correction $e$ neglected, it depends only on the length $l$ of the tube and on which overtone sounds. So it does not change with the temperature of the air.

The third quote links speed, note and wavelength: $v = f \lambda$. With $\lambda$ fixed, the note goes with the speed.

The fourth and fifth quotes say that the speed of sound goes up with the temperature, and give the ratio of the speeds. The sixth quote shows how to use it: the temperatures are written on the kelvin scale, with $0\ ^\circ\text{C}$ as $273\ \text{K}$.

The last quote is about beats: the number of loud sounds in one second.

For example, the pipes of a church organ. On a hot afternoon the pipes sound a little higher, because the pipes stay the same length and sound moves faster in the warm air.

Required equations

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

$\dfrac{f_2}{f_1} = \sqrt{\dfrac{T_2}{T_1}}$

$f_1 = f - 4, \quad f_2 = f + 4$

$\dfrac{f + 4}{f - 4} = \sqrt{\dfrac{T_2}{T_1}}$

Applying it to this question

The ratio $\sqrt{\dfrac{T_2}{T_1}}$ has no unit. The frequencies are in $\text{Hz}$ and the beat frequency is in $\text{Hz}$, so $f$ comes out in $\text{Hz}$.

We use $T_2 > T_1$. So $f_2 > f_1$, and the $4\ \text{Hz}$ is taken away from $f$ for the cooler tube and added to $f$ for the hotter one.

$324$

$T_1 = 51 + 273$

$400$

$T_2 = 127 + 273$

$1.11$

$\dfrac{f_2}{f_1} = \sqrt{\dfrac{400}{324}} = \dfrac{20}{18}$

$\dfrac{f + 4}{f - 4} = \dfrac{20}{18}$

$18(f + 4) = 20(f - 4)$

$18f + 72 = 20f - 80$

$76$

$f = \dfrac{152}{2}$

Reasoning and calculation

So $f = 76\ \text{Hz}$. The tube notes are $f_1 = 72\ \text{Hz}$ and $f_2 = 80\ \text{Hz}$.

Their ratio is $\dfrac{80}{72} = \dfrac{20}{18}$, which is what the temperatures need. That is option (4).

Checks

$4 * (r + 1) / (r - 1)$

Does it make sense?

The fork is $76\ \text{Hz}$. The cooler tube sounds $72\ \text{Hz}$, which is $4\ \text{Hz}$ below the fork. The hotter tube sounds $80\ \text{Hz}$, which is $4\ \text{Hz}$ above the fork. Both give $4\ \text{Hz}$ beats, as the question says.

The hotter tube has the higher note, as it should, because sound moves faster in the warmer air.

The temperature ratio is $\dfrac{400}{324}$ and the note ratio is $\dfrac{80}{72}$. The note ratio is the square root of the temperature ratio, as the speed formula says.

Conclusion

$76\ \text{Hz}$

The tube notes are $72\ \text{Hz}$ and $80\ \text{Hz}$, so the fork is $76\ \text{Hz}$.

The answer is (4).

For the same tube and the same overtone, with the end correction neglected, the wavelength stays the same, so the note goes with the speed of sound, which goes with $\sqrt{T}$ in kelvin.

If the fork gives the same beats with two different notes, it sits half way between them.

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